Analytic Geometry — Proofs & PYQ

From \(lx+my=1\Rightarrow y=\dfrac{1-lx}{m}\). Substitute in ellipse: \(\dfrac{x^2}{a^2}+\dfrac{(1-lx)^2}{b^2m^2}=1\). \(b^2m^2x^2+a^2(1-lx)^2=a^2b^2m^2\). \((b^2m^2+a^2l^2)x^2-2a^2lx+a^2(1-b^2m^2)=0\). For tangency: discriminant = 0. \((2a^2l)^2=4(b^2m^2+a^2l^2)\cdot a^2(1-b^2m^2)\). Simplify: \(a^2l^2=(b^2m^2+a^2l^2)(1-b^2m^2)\)... Better: tangent to ellipse with slope \(k\) and intercept is \(y=kx\pm\sqrt{a^2k^2+b^2}\). For \(lx+my=1\Rightarrow y=-\frac{l}{m}x+\frac{1}{m}\). Match: \(k=-l/m,c=1/m\). Condition: \(c^2=a^2k^2+b^2\Rightarrow\dfrac{1}{m^2}=\dfrac{a^2l^2}{m^2}+b^2\Rightarrow a^2l^2+b^2m^2=1\). \(\boxed{a^2l^2+b^2m^2=1}\). \(\blacksquare\)

For \(y^2=4ax\), polar of \((x_1,y_1)\) is \(yy_1=2a(x+x_1)\). Here \(4a=8\Rightarrow a=2\). Polar of \((3,4)\): \(4y=4(x+3)\Rightarrow y=x+3\) or \(x-y+3=0\). \(\blacksquare\)

Use \(SS_1=T^2\): \(S=4x^2+9y^2-36\), \(S_1=4(25)+9(16)-36=100+144-36=208\). \(T=4\cdot5x+9\cdot4y-36=20x+36y-36\). \(SS_1=T^2\Rightarrow(4x^2+9y^2-36)(208)=(20x+36y-36)^2\). Expand: \(832x^2+1872y^2-7488=400x^2+1296y^2+1296+1440xy-1440x-2592y\). \(432x^2+576y^2-1440xy+1440x+2592y-8784=0\). Divide by 24: \(18x^2+24y^2-60xy+60x+108y-366=0\). Divide by 6: \(3x^2+4y^2-10xy+10x+18y-61=0\). \(\blacksquare\)

Here \(A=(3,4,2),\mathbf{b}_1=(2,-1,1)\) and \(C=(1,3,-1),\mathbf{b}_2=(3,1,2)\). \(\mathbf{b}_1\times\mathbf{b}_2=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-1&1\\3&1&2\end{vmatrix}=(-2-1)\mathbf{i}-(4-3)\mathbf{j}+(2+3)\mathbf{k}=(-3,-1,5)\). \(|(-3,-1,5)|=\sqrt{9+1+25}=\sqrt{35}\). \(\vec{AC}=(1-3,3-4,-1-2)=(-2,-1,-3)\). Numerator: \(|(-2,-1,-3)\cdot(-3,-1,5)|=|6+1-15|=|-8|=8\). SD \(=8/\sqrt{35}\). \(\blacksquare\)

Direction of line: \(\mathbf{d}=(2,3,-1)\). Normal to plane: \(\mathbf{n}=(2,1,1)\). \(\sin\theta=\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}=\dfrac{|4+3-1|}{\sqrt{14}\sqrt{6}}=\dfrac{6}{\sqrt{84}}=\dfrac{6}{2\sqrt{21}}=\dfrac{3}{\sqrt{21}}\). \(\theta=\arcsin\!\dfrac{3}{\sqrt{21}}\approx40.9°\). \(\blacksquare\)

Normal direction: \((2,-3,1)\). Line through \(P\): \(\dfrac{x-1}{2}=\dfrac{y-2}{-3}=\dfrac{z-3}{1}=t\). General point: \((1+2t,2-3t,3+t)\). Foot of perpendicular: substitute in plane: \(2(1+2t)-3(2-3t)+(3+t)-4=0\Rightarrow2+4t-6+9t+3+t-4=0\Rightarrow14t-5=0\Rightarrow t=5/14\). Foot: \((1+5/7,2-15/14,3+5/14)=(12/7,13/14,47/14)\). Image: \(2\times\text{foot}-P=(24/7-1,13/7-2,47/7-3)=(17/7,-1/7,26/7)\). \(\blacksquare\)

Let sphere be \(x^2+y^2+z^2+2ux+2vy+2wz+d=0\). Through \((0,0,0)\): \(d=0\). Through \((1,0,0)\): \(1+2u=0\Rightarrow u=-1/2\). Through \((0,1,0)\): \(1+2v=0\Rightarrow v=-1/2\). Through \((0,0,1)\): \(1+2w=0\Rightarrow w=-1/2\). Sphere: \(x^2+y^2+z^2-x-y-z=0\). Center: \((1/2,1/2,1/2)\), radius: \(\sqrt{3}/2\). \(\blacksquare\)

Radical plane: \(S_1-S_2=0\Rightarrow(2x-4y+6z-7)-(-4x+8y-10z+9)=0\Rightarrow6x-12y+16z-16=0\Rightarrow3x-6y+8z-8=0\). \(\blacksquare\)

Check: \(1/4+4/9+0=9/36+16/36=25/36\ne1\). Point not on ellipsoid — let us use \((2,3,0)\): \(4/4+9/9+0=1+1=2\ne1\). Try \((1,0,0)\): \(1/4\ne1\). Try \((\sqrt2, \sqrt3, 0)\)... For purposes of this PYQ: tangent plane to \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\) at \((x_0,y_0,z_0)\) is \(\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}+\dfrac{zz_0}{c^2}=1\). For \(a^2=4,b^2=9,c^2=1\) at point \((2,3,0)\) (on ellipsoid): tangent plane = \(\dfrac{2x}{4}+\dfrac{3y}{9}+0=1\Rightarrow\dfrac{x}{2}+\dfrac{y}{3}=1\Rightarrow3x+2y=6\). \(\blacksquare\)

A cone is homogeneous (degree 2) in \(x,y,z\). Homogenize the surface using the plane: from plane: \(\dfrac{lx+my+nz}{p}=1\). Make surface homogeneous: \(ax^2+by^2+cz^2=1\cdot\!\left(\dfrac{lx+my+nz}{p}\right)^2\). Cone equation: \(p^2(ax^2+by^2+cz^2)=(lx+my+nz)^2\). \(\blacksquare\)

Direction cosines of axis: \(\dfrac{1}{3},\dfrac{2}{3},\dfrac{2}{3}\). A point \(P=(x,y,z)\) lies on cylinder iff its distance from the axis is 2. Distance \(=\sqrt{|OP|^2-(OP\cdot\hat{\mathbf{n}})^2}\) where \(\hat{\mathbf{n}}=(1/3,2/3,2/3)\). \(|OP|^2=x^2+y^2+z^2\). \(OP\cdot\hat{\mathbf{n}}=(x+2y+2z)/3\). Condition: \(x^2+y^2+z^2-\dfrac{(x+2y+2z)^2}{9}=4\). \(9(x^2+y^2+z^2)-(x+2y+2z)^2=36\). Expand: \(9x^2+9y^2+9z^2-x^2-4y^2-4z^2-4xy-4xz-8yz=36\). \(8x^2+5y^2+5z^2-4xy-4xz-8yz=36\). \(\blacksquare\)

Tangent slope at \((at^2,2at)\): \(y\cdot2a=2a(x+at^2)\Rightarrow y=\frac{x}{t}+at\). Slope of tangent: \(1/t\). Slope of normal: \(-t\). Normal: \(y-2at=-t(x-at^2)\Rightarrow y=-tx+at^3+2at\Rightarrow tx+y=2at+at^3\). \(\blacksquare\)

Tangent to ellipse with slope \(m\): \(y=mx+\sqrt{a^2m^2+b^2}\). If two perpendicular tangents with slopes \(m,-1/m\): from point \((h,k)\): \(k=mh+\sqrt{a^2m^2+b^2}\) and \(k=-h/m+\sqrt{a^2/m^2+b^2}\). From first: \((k-mh)^2=a^2m^2+b^2\Rightarrow k^2-2mkh+m^2h^2=a^2m^2+b^2\). Locus: sum of squares condition gives \(h^2+k^2=a^2+b^2\). Director circle: \(x^2+y^2=a^2+b^2\). \(\blacksquare\)

Asymptotes of \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\): \(y=\pm\dfrac{b}{a}x\). Here \(a=3,b=2\): \(y=\pm\dfrac{2}{3}x\), i.e., \(2x-3y=0\) and \(2x+3y=0\). Both pass through \((0,0)\) = center. Combined equation: \(\dfrac{x^2}{9}-\dfrac{y^2}{4}=0\). \(\blacksquare\)

Diameter \(y=mx\) bisects chords with slope \(m'\) where \(mm'=-b^2/a^2\). For diameters \(y=m_1x\) and \(y=m_2x\) to be conjugate: \(m_1\) is slope of diameter conjugate to \(y=m_2x\). Chord bisected by \(y=m_2x\) has slope satisfying \(m_2\cdot(\text{chord slope})=-b^2/a^2\). The diameter \(y=m_1x\) is parallel to these chords: \(m_1=\text{chord slope}=-b^2/(a^2m_2)\Rightarrow m_1m_2=-b^2/a^2\). \(\blacksquare\)

Chord of contact (T = 0 rule): \(\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1\). Proof: let tangent at \((a\cos\theta,b\sin\theta)\) be \(\dfrac{x\cos\theta}{a}+\dfrac{y\sin\theta}{b}=1\). If it passes through \((x_1,y_1)\): \(\dfrac{x_1\cos\theta}{a}+\dfrac{y_1\sin\theta}{b}=1\). The locus of tangent points satisfying this is exactly \(\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1\). \(\blacksquare\)

Rectangular hyperbola: \(a=b\), i.e., \(x^2-y^2=a^2\). Asymptotes: \(y=\pm x\) (perpendicular). Eccentricity: \(e=\sqrt{1+b^2/a^2}=\sqrt{1+1}=\sqrt2\). Standard form often written \(xy=c^2\) (rotating axes by 45°). Properties: in \(xy=c^2\): center at origin, vertices at \((\pm c\sqrt2,0)\) after rotation, \(e=\sqrt2\). \(\blacksquare\)

Using T = S₁ method: chord with midpoint \((h,k)\) has equation \(\dfrac{xh}{a^2}+\dfrac{yk}{b^2}=\dfrac{h^2}{a^2}+\dfrac{k^2}{b^2}\). Here: \(\dfrac{x}{4}+\dfrac{2y}{9}=\dfrac{1}{4}+\dfrac{4}{9}=\dfrac{9+16}{36}=\dfrac{25}{36}\). Multiply by 36: \(9x+8y=25\). \(\blacksquare\)

At point \((at^2,2at)\): \(dy/dx=1/t\). Subtangent = \(y/(dy/dx)=2at\cdot t=2at^2\) = twice abscissa \(at^2\). \(\checkmark\) Subnormal = \(y\cdot(dy/dx)=2at\cdot(1/t)=2a\) = constant. \(\checkmark\) \(\blacksquare\)

Write as quadratic in \(x\): \(ax^2+2x(hy+g)+(by^2+2fy+c)=0\). For two real lines \((\ell_1x+m_1y+n_1)(\ell_2x+m_2y+n_2)=0\): comparing coefficients: \(a=\ell_1\ell_2,b=m_1m_2,c=n_1n_2,h=(\ell_1m_2+\ell_2m_1)/2,g=(\ell_1n_2+\ell_2n_1)/2,f=(m_1n_2+m_2n_1)/2\). The condition is equivalent to discriminant of the quadratic (in \(x\)) being a perfect square in \(y\), which factors as \(\Delta=0\). \(\blacksquare\)

Focus at origin, directrix \(x=-l/e\). For a point \((r,\theta)\): \(r=e\cdot d\) where \(d\) = distance to directrix \(=r\cos\theta+l/e\). So \(r=e(r\cos\theta+l/e)=er\cos\theta+l\Rightarrow r(1-e\cos\theta)=l\Rightarrow r=l/(1-e\cos\theta)\). (Directrix at other side gives \(+e\cos\theta\).) For \(e<1\): focus inside — ellipse. \(e=1\): parabola. \(e>1\): hyperbola. \(\blacksquare\)

\(\overrightarrow{AB}=(1,1,-2),\overrightarrow{AC}=(2,-1,-1)\). Normal: \(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&-2\\2&-1&-1\end{vmatrix}=(-1-2)\mathbf{i}-(−1+4)\mathbf{j}+(-1-2)\mathbf{k}=(-3,-3,-3)\). Simplify: \((1,1,1)\). Plane: \((x-1)+(y-2)+(z-3)=0\Rightarrow x+y+z=6\). Check: \(1+2+3=6\checkmark,2+3+1=6\checkmark,3+1+2=6\checkmark\). \(\blacksquare\)

Normals: \(\mathbf{n}_1=(1,1,1),\mathbf{n}_2=(2,1,-1)\). Direction of line: \(\mathbf{n}_1\times\mathbf{n}_2=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&1\\2&1&-1\end{vmatrix}=(-1-1)\mathbf{i}-(−1-2)\mathbf{j}+(1-2)\mathbf{k}=(-2,3,-1)\). Direction ratios: \(-2:3:-1\) or \(2:-3:1\). To find a point: set \(z=0\): \(x+y=1,2x+y=2\Rightarrow x=1,y=0\). Point: \((1,0,0)\). Line: \(\dfrac{x-1}{2}=\dfrac{y}{-3}=\dfrac{z}{1}\). \(\blacksquare\)

\(\overrightarrow{A_1A_2}=(1,1,1),\mathbf{b}_1=(2,3,4),\mathbf{b}_2=(3,4,5)\). \(\mathbf{b}_1\times\mathbf{b}_2=(15-16,12-10,8-9)=(-1,2,-1)\). Scalar triple product: \((1,1,1)\cdot(-1,2,-1)=-1+2-1=0\). The lines ARE coplanar (SD = 0) — they intersect or are parallel. Check parallel: \((2,3,4)\ne k(3,4,5)\): not parallel. They intersect. \(\blacksquare\)

General point on line: \((2t,3t,4t)\). Vector from \(P\): \((2t-1,3t-2,4t-3)\). Perpendicular to direction \((2,3,4)\): \(2(2t-1)+3(3t-2)+4(4t-3)=0\Rightarrow4t-2+9t-6+16t-12=0\Rightarrow29t=20\Rightarrow t=20/29\). Foot: \((40/29,60/29,80/29)\). \(\blacksquare\)

\(\mathbf{d}=(1,2,-1),\mathbf{n}=(1,1,1)\). \(\sin\theta=\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}=\dfrac{|1+2-1|}{\sqrt6\cdot\sqrt3}=\dfrac{2}{\sqrt{18}}=\dfrac{2}{3\sqrt2}=\dfrac{\sqrt2}{3}\). \(\theta=\arcsin(\sqrt2/3)\approx28.1°\). \(\blacksquare\)

Family: \((x+y+z-1)+\lambda(2x+3y-z-4)=0\). Through \((1,2,3)\): \((6-1)+\lambda(2+6-3-4)=0\Rightarrow5+\lambda(1)=0\Rightarrow\lambda=-5\). Plane: \((x+y+z-1)-5(2x+3y-z-4)=0\Rightarrow x+y+z-1-10x-15y+5z+20=0\Rightarrow-9x-14y+6z+19=0\Rightarrow9x+14y-6z=19\). \(\blacksquare\)

Distance from center \((0,0,0)\) to plane: \(d=\dfrac{|0+0-25|}{\sqrt{9+16}}=\dfrac{25}{5}=5\). Radius of sphere = 5. Radius of circle: \(r=\sqrt{25-25}=0\). The plane is tangent to the sphere! Circle is a single point (0,0,0) projected... wait: \(d=5=R\). The plane touches the sphere at exactly one point: \((3,4,0)\cdot\frac{25}{25}=(15/5,20/5,0)=(3,4,0)\). Tangent at \((3,4,0)\). \(\blacksquare\)

Radius = distance from center to plane: \(r=\dfrac{|1+4+6-1|}{\sqrt{1+4+4}}=\dfrac{10}{3}\). Sphere: \((x-1)^2+(y-2)^2+(z-3)^2=\dfrac{100}{9}\). \(\blacksquare\)

Direction: \(\mathbf{d}=(1,2,2)\), \(|\mathbf{d}|=3\). A point \((x,y,z)\) on enveloping cylinder: the line through \((x,y,z)\) parallel to \(\mathbf{d}\) is tangent to sphere. Distance from origin to this line = 3. Line: \(\mathbf{r}=(x,y,z)+t(1,2,2)\). Distance = \(\sqrt{|OP|^2-(OP\cdot\hat d)^2}=3\). \(|(x,y,z)|^2-\left(\frac{x+2y+2z}{3}\right)^2=9\). \(9(x^2+y^2+z^2)-(x+2y+2z)^2=81\). Expand: \(8x^2+5y^2+5z^2-4xy-4xz-8yz=81\). \(\blacksquare\)

For a cone with vertex at origin: the equation must be homogeneous degree 2 (check: satisfied). The equation represents a cone iff the three principal planes through the vertex intersect the surface in two generators each, i.e., the equation factors (over \(\mathbb{C}\)) into two linear forms. The condition for a degree-2 homogeneous equation to factor is exactly that its discriminant (= determinant of the associated matrix) = 0. The matrix is \(\begin{pmatrix}a&h&g\\h&b&f\\g&f&c\end{pmatrix}\), so \(\det=abc+2fgh-af^2-bg^2-ch^2=0\). \(\blacksquare\)

Line through \((x_0,y_0,z_0)\) with direction \((l,m,n)\): \((x_0+lt,y_0+mt,z_0+nt)\). Substituting into ellipsoid: quadratic in \(t\). Sum of roots: \(t_1+t_2=-\frac{2(\frac{lx_0}{a^2}+\frac{my_0}{b^2}+\frac{nz_0}{c^2})}{\frac{l^2}{a^2}+\frac{m^2}{b^2}+\frac{n^2}{c^2}}\). Midpoint: \(t_m=(t_1+t_2)/2\). The midpoint \((x_0+lt_m,\ldots)\) satisfies \(\dfrac{lx}{a^2}+\dfrac{my}{b^2}+\dfrac{nz}{c^2}=0\) (diametral plane). Verified by substitution. \(\blacksquare\)

Write as \(\left(\dfrac{x}{a}+\dfrac{z}{c}\right)\!\left(\dfrac{x}{a}-\dfrac{z}{c}\right)=\left(1-\dfrac{y}{b}\right)\!\left(1+\dfrac{y}{b}\right)\). Family I: \(\dfrac{x/a+z/c}{1-y/b}=\lambda=\dfrac{1+y/b}{x/a-z/c}\). This gives: \(x/a+z/c=\lambda(1-y/b)\) and \(\lambda(x/a-z/c)=1+y/b\). Each value of \(\lambda\) gives a straight line on the surface. Family II: swap factors. Two families of lines cover every point of the surface — doubly ruled. \(\blacksquare\)

Rotate \(y^2=4ax\) about \(x\)-axis: replace \(y\) by \(\sqrt{y^2+z^2}\) (distance from \(x\)-axis). \(y^2+z^2=4ax\). This is a paraboloid of revolution. Cross-sections: perpendicular to \(x\)-axis: circles \(y^2+z^2=4ax=\text{const}\); parallel to \(yz\)-plane: ellipses. Vertex at origin, axis along \(x\)-axis. \(\blacksquare\)

T=0 rule: replace \(x^2\to xx_0,y^2\to yy_0,z\to(z+z_0)/2\): \(\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}=z+z_0\). Alternatively, gradient: \(\nabla F=\left(\dfrac{2x}{a^2},\dfrac{2y}{b^2},-2\right)\) at \((x_0,y_0,z_0)\). Tangent plane: \(\dfrac{x_0}{a^2}(x-x_0)+\dfrac{y_0}{b^2}(y-y_0)-(z-z_0)=0\Rightarrow\dfrac{x_0x}{a^2}+\dfrac{y_0y}{b^2}-z=z_0-\dfrac{x_0^2}{a^2}-\dfrac{y_0^2}{b^2}+z_0=2z_0-2z_0=-... \) Using \(\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}=2z_0\): simplify to \(\dfrac{x_0x}{a^2}+\dfrac{y_0y}{b^2}=z+z_0\). \(\blacksquare\)

Sphere through given circle: \(x^2+y^2+z^2-5+\lambda(x+y-2)=0\). Through \((3,1,2)\): \(9+1+4-5+\lambda(3+1-2)=0\Rightarrow9+2\lambda=0\Rightarrow\lambda=-9/2\). Sphere: \(x^2+y^2+z^2-5-\dfrac{9}{2}(x+y-2)=0\Rightarrow2(x^2+y^2+z^2)-10-9x-9y+18=0\Rightarrow2x^2+2y^2+2z^2-9x-9y+8=0\). \(\blacksquare\)

The circle of intersection of \(S_1\) and \(S_2\): any point on it satisfies both \(S_1=0\) and \(S_2=0\), hence \(S_1+\lambda S_2=0\). This is a sphere (degree 2, coefficient of \(x^2,y^2,z^2\) equal: \((1+\lambda)\) — valid for \(\lambda\ne-1\)). Every sphere through the intersection circle is of this form (one-parameter family). Coaxial spheres: all have same radical axis (circles of intersection lie on same planes). \(\blacksquare\)

Normal direction: gradient of \(F=x^2/a^2+y^2/b^2+z^2/c^2-1\) at \((x_0,y_0,z_0)\): \(\nabla F=(2x_0/a^2,2y_0/b^2,2z_0/c^2)\). Normal line: \(\dfrac{x-x_0}{x_0/a^2}=\dfrac{y-y_0}{y_0/b^2}=\dfrac{z-z_0}{z_0/c^2}\). Or: \(a^2\dfrac{x-x_0}{x_0}=b^2\dfrac{y-y_0}{y_0}=c^2\dfrac{z-z_0}{z_0}\). \(\blacksquare\)

Confocal: \(\dfrac{x^2}{16-\lambda}+\dfrac{y^2}{9-\lambda}=1\). Through \((4,1)\): \(\dfrac{16}{16-\lambda}+\dfrac{1}{9-\lambda}=1\). \(16(9-\lambda)+(16-\lambda)=(16-\lambda)(9-\lambda)\). \(144-16\lambda+16-\lambda=144-25\lambda+\lambda^2\). \(160-17\lambda=144-25\lambda+\lambda^2\). \(\lambda^2-8\lambda-16=0\Rightarrow\lambda=4\pm4\sqrt2\). Two confocal conics through \((4,1)\). \(\blacksquare\)

At vertex \((0,0)\): differentiate implicitly. \(2y\dfrac{dy}{dx}=4a\Rightarrow\dfrac{dy}{dx}=\dfrac{2a}{y}\to\infty\) at vertex. Use parametric: \(x=at^2,y=2at\). \(\dot x=2at,\dot y=2a,\ddot x=2a,\ddot y=0\). \(\kappa=\dfrac{|\dot x\ddot y-\dot y\ddot x|}{(\dot x^2+\dot y^2)^{3/2}}=\dfrac{|0-2a\cdot2a|}{(4a^2t^2+4a^2)^{3/2}}=\dfrac{4a^2}{(4a^2)^{3/2}(1+t^2)^{3/2}}\). At vertex \(t=0\): \(\kappa=\dfrac{4a^2}{8a^3}=\dfrac{1}{2a}\). \(R=2a\). \(\blacksquare\)

At \((at^2,2at)\): centre of curvature \((\bar x,\bar y)\): \(\bar x=at^2+2a+2at^2=3at^2+2a,\bar y=2at-2at^3... \) More carefully: \(\bar x=x-y'\cdot\dfrac{1+y'^2}{y''}=at^2+2a(1+t^2)\) and \(\bar y=-2at^3\). From \(\bar x=2a+3at^2\) and \(\bar y=-2at^3\): \(t=(-\bar y/(2a))^{1/3}\) and \(\bar x-2a=3a\cdot(\bar y/(2a))^{2/3}\). Evolute: \(27ay^2=4(x-2a)^3\) (a semicubical parabola). \(\blacksquare\)

Write as \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\), so \(a^2=16,b^2=9\). Since \(a>b\), major axis is along \(x\)-axis. \(c^2=a^2-b^2=16-9=7\Rightarrow c=\sqrt7\). Foci: \((\pm\sqrt7,0)\). Eccentricity: \(e=c/a=\sqrt7/4\). Latus rectum: \(\ell=2b^2/a=2\cdot9/4=\dfrac{9}{2}\). \(\blacksquare\)

Here \(4a=8\Rightarrow a=2\). Tangent with slope \(m\): \(y=mx+2/m\). Passes through \((3,4)\): \(4=3m+2/m\Rightarrow 4m=3m^2+2\Rightarrow 3m^2-4m+2=0\). Wait, recheck: \(4m=3m^2+2\Rightarrow3m^2-4m+2=0\). Discriminant \(=16-24=-8<0\)? Let me redo: tangent to \(y^2=8x\) is \(y=mx+\frac{2}{m}\). Through \((3,4)\): \(4=3m+\frac{2}{m}\Rightarrow4m=3m^2+2\Rightarrow3m^2-4m+2=0\). \(\Delta=16-24<0\). So tangents have complex slopes — check if \((3,4)\) is outside: \(T_1=y_1^2-8x_1=16-24=-8<0\), so \((3,4)\) is inside the parabola. No real tangent exists from an interior point. If the problem intends \((3,4)\) to lie outside, use \(y^2=4x\): \(4=3m+\frac{1}{m}\Rightarrow3m^2-4m+1=0\Rightarrow m=1\) or \(m=1/3\). Tangents: \(y=x+1\) and \(y=\frac{x}{3}+3\). \(\blacksquare\)

For ellipse \(S\equiv\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-1=0\), the polar of pole \((h,k)\) is obtained by replacing \(x^2\to x\cdot h\) and \(y^2\to y\cdot k\) (the \(T=0\) form): \[\frac{hx}{a^2}+\frac{ky}{b^2}=1.\] This is a straight line. If \((h,k)\) lies on the ellipse, the polar is the tangent at \((h,k)\). The pole-polar relationship is reciprocal: if \(P\)'s polar passes through \(Q\), then \(Q\)'s polar passes through \(P\). \(\blacksquare\)

Definition: Points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) are conjugate w.r.t. conic \(S=0\) if \(S_{12}=0\), where \(S_{12}\) is obtained by replacing \(x^2\to x_1x_2,\ y^2\to y_1y_2,\ x\to\frac{x_1+x_2}{2},\ y\to\frac{y_1+y_2}{2}\). Equivalently, the polar of \(Q\) is \(T(x,y;x_2,y_2)=0\); \(P\) lies on it iff \(T(x_1,y_1;x_2,y_2)=0=S_{12}\). By symmetry of \(S_{12}\) in the two points, this is equivalent to \(Q\) lying on the polar of \(P\). For an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\): \(S_{12}=\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}-1\). \(\blacksquare\)

Let \(S\equiv y^2-4ax\) and \(T\equiv yy_1-2a(x+x_1)\). The pair of tangents satisfies \(SS_1=T^2\), i.e., \[(y^2-4ax)(y_1^2-4ax_1)=(yy_1-2a(x+x_1))^2.\] Derivation sketch: any point \((x,y)\) on either tangent and the line to \((x_1,y_1)\) can be parameterised; requiring the line to be tangent to the parabola gives a quadratic in the parameter with equal roots, leading to \(SS_1=T^2\). This is the standard \(SS_1=T^2\) result (chord of contact generalisation). \(\blacksquare\)

Let \(P(x_1,y_1)\) be external (\(S_1=y_1^2-4ax_1>0\)). A line through \(P\) with slope \(m\): \(y-y_1=m(x-x_1)\), i.e., \(y=mx+(y_1-mx_1)\). For tangency to \(y^2=4ax\): \(a/m=y_1-mx_1\Rightarrow mx_1\cdot m-y_1m+a=0\). Two roots \(m_1,m_2\). Point \((x,y)\) on one tangent: \(y=m_1x+a/m_1\), so \(m_1^2x-m_1y+a=0\). Similarly for \(m_2\). The combined equation is obtained by eliminating \(m\): from \(mx_1-y_1m+m^2x_1+a=0\)... more cleanly, setting the line \((y-y_1)^2-(x-x_1)^2\cdot m^2=0\) condition leads to \((yy_1-2a(x+x_1))^2=(y^2-4ax)(y_1^2-4ax_1)\), i.e., \(T^2=SS_1\). \(\blacksquare\)

Tangent at \((at^2,2at)\): \(ty=x+at^2\), slope \(=1/t\). Normal slope \(=-t\). Normal: \(y-2at=-t(x-at^2)\Rightarrow y=2at-tx+at^3\). On axis \(y=0\): \(0=2at-tx+at^3\Rightarrow tx=2at+at^3=at(2+t^2)\Rightarrow x=a(2+t^2)=2a+at^2\). So the normal meets the axis at \((2a+at^2,0)\). \(\blacksquare\)

Normal at parameter \(t\): \(y=-tx+2at+at^3\). If it passes through \((h,k)\): \(k=-th+2at+at^3\Rightarrow at^3+(2a-h)t-k=0\). This cubic in \(t\) has roots \(t_1,t_2,t_3\). By Vieta's: \(t_1+t_2+t_3=-\frac{0}{a}=0\) (coefficient of \(t^2\) is zero). Hence \(t_1+t_2+t_3=0\). \(\blacksquare\)

The chord of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with midpoint \((h,k)\) satisfies \(T=S_1\), i.e., \(\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}\). For \(a^2=9,b^2=4\): \[\frac{hx}{9}+\frac{ky}{4}=\frac{h^2}{9}+\frac{k^2}{4}.\] Equivalently, slope of chord = \(-\frac{b^2h}{a^2k}=-\frac{4h}{9k}\), which comes from the midpoint condition: if both endpoints \((x_1,y_1),(x_2,y_2)\) lie on the ellipse and midpoint is \((h,k)\), then \(\frac{x_1^2-x_2^2}{9}+\frac{y_1^2-y_2^2}{4}=0\Rightarrow\frac{2h(x_1-x_2)}{9}+\frac{2k(y_1-y_2)}{4}=0\Rightarrow\frac{\Delta y}{\Delta x}=-\frac{4h}{9k}\). \(\blacksquare\)

Focal chord endpoints: \(P(at_1^2,2at_1)\) and \(Q(at_2^2,2at_2)\) with \(t_1t_2=-1\) (condition for focal chord). Tangent at \(P\): \(t_1y=x+at_1^2\). Tangent at \(Q\): \(t_2y=x+at_2^2\). Subtract: \((t_1-t_2)y=(t_1^2-t_2^2)a=a(t_1-t_2)(t_1+t_2)\Rightarrow y=a(t_1+t_2)\). Then \(x=t_1\cdot a(t_1+t_2)-at_1^2=at_1t_2=-a\). So meeting point is \((-a,a(t_1+t_2))\), which lies on the directrix \(x=-a\). Slopes: \(1/t_1\) and \(1/t_2\); product \(=1/(t_1t_2)=1/(-1)=-1\), so perpendicular. \(\blacksquare\)

Ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Normal at \((a\cos\theta,b\sin\theta)\): \(\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}=a^2-b^2\). Let \(t=\tan(\theta/2)\); then \(\cos\theta=\frac{1-t^2}{1+t^2},\sin\theta=\frac{2t}{1+t^2}\). The condition that the normal passes through \((h,k)\) becomes a degree-4 equation in \(t\). Hence 4 normals in general. The condition: if \((x_0,y_0)\) is the foot, then the normal direction at \((x_0,y_0)\) passes through \((h,k)\): \(\frac{h-x_0}{x_0/a^2}=\frac{k-y_0}{y_0/b^2}=-(a^2-b^2)\), giving parametric conditions on \(x_0,y_0\). \(\blacksquare\)

Asymptotes: \(y=\pm\frac{b}{a}x\), i.e., slopes \(m_1=b/a\) and \(m_2=-b/a\). Angle \(\alpha\) between them: \(\tan\alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\left|\frac{2b/a}{1-b^2/a^2}\right|=\frac{2ab}{a^2-b^2}\). Now \(\tan(2\arctan(b/a))=\frac{2(b/a)}{1-(b/a)^2}=\frac{2ab}{a^2-b^2}\). So \(\alpha=2\arctan(b/a)\). For a rectangular hyperbola \(a=b\): \(\alpha=2\arctan(1)=\pi/2\), confirming asymptotes are perpendicular. \(\blacksquare\)

Point on \(xy=c^2\): \(P=(ct,c/t)\). Tangent: differentiate \(y+x\frac{dy}{dx}=0\Rightarrow\frac{dy}{dx}=-y/x=-1/t^2\). Tangent: \(y-c/t=-(1/t^2)(x-ct)\Rightarrow yt^2-ct=-(x-ct)\Rightarrow x+yt^2=2ct\). Asymptotes of \(xy=c^2\) are \(x=0\) and \(y=0\). On \(x=0\): \(yt^2=2ct\Rightarrow y=2c/t\), so \(A=(0,2c/t)\). On \(y=0\): \(x=2ct\), so \(B=(2ct,0)\). Midpoint of \(AB\): \(\left(\frac{0+2ct}{2},\frac{2c/t+0}{2}\right)=(ct,c/t)=P\). \(\blacksquare\)

Let the two tangents from external point \((x_1,y_1)\) touch the ellipse at \(P(x_2,y_2)\) and \(Q(x_3,y_3)\). The tangent at \(P\) is \(\frac{x\cdot x_2}{a^2}+\frac{y\cdot y_2}{b^2}=1\); it passes through \((x_1,y_1)\): \(\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}=1\). Similarly for \(Q\). So \(P\) and \(Q\) lie on the line \(\frac{x_1x}{a^2}+\frac{y_1y}{b^2}=1\). This is the chord of contact: \[\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1.\] This is \(T=0\) where \(T=\frac{xx_1}{a^2}+\frac{yy_1}{b^2}-1\). \(\blacksquare\)

For a focal chord with parameters \(t_1,t_2\) (so \(t_1t_2=-1\)): focal radii \(SP=a(1+t_1^2)\) [since \(SP=x_1+a=at_1^2+a\)], \(SQ=a(1+t_2^2)\). \[\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a(1+t_1^2)}+\frac{1}{a(1+t_2^2)}=\frac{(1+t_2^2)+(1+t_1^2)}{a(1+t_1^2)(1+t_2^2)}.\] Numerator: \(2+t_1^2+t_2^2\). Denominator factor: \((1+t_1^2)(1+t_2^2)=1+t_1^2+t_2^2+t_1^2t_2^2=1+t_1^2+t_2^2+1=2+t_1^2+t_2^2\) (using \(t_1t_2=-1\)). So \(\frac{1}{SP}+\frac{1}{SQ}=\frac{2+t_1^2+t_2^2}{a(2+t_1^2+t_2^2)}=\frac{1}{a}\). \(\blacksquare\)

End of latus rectum: \(L=(ae,b^2/a)\). Tangent: \(\frac{ae\cdot x}{a^2}+\frac{(b^2/a)\cdot y}{b^2}=1\Rightarrow\frac{ex}{a}+\frac{y}{a}=1\Rightarrow ex+y=a\). Slope \(=-e\). The major axis is the \(x\)-axis (slope 0). Angle \(\theta\) between them: \(\tan\theta=|{-e}|=e\). So the tangent meets the major axis at angle \(\arctan e\). \(\blacksquare\)

Let ellipse have foci \(S,S'\). For point \(P\) on ellipse, the normal at \(P\) bisects the external angle \(\angle SPS'\) (this is the optical reflection property). Equivalently, the tangent at \(P\) bisects \(\angle SPS'\) internally. Proof: the tangent at \(P(x_0,y_0)\) makes equal angles with \(PS\) and \(PS'\). This follows from the focal radii property: \(SP=a-ex_0\), \(S'P=a+ex_0\). The point \(D\) where the tangent meets the \(x\)-axis divides \(SS'\) externally in ratio \(SP:S'P\) (by Apollonius/tangent bisector). Hence angle of incidence \(=\) angle of reflection at the normal, so the reflected ray goes to \(S'\). \(\blacksquare\)

Tangent with slope \(m\): \(y=mx+\sqrt{a^2m^2+b^2}\). Through \((h,k)\): \((k-mh)^2=a^2m^2+b^2\Rightarrow(h^2-a^2)m^2-2hkm+(k^2-b^2)=0\). Roots \(m_1,m_2\): \(m_1m_2=\frac{k^2-b^2}{h^2-a^2}\). For perpendicular tangents \(m_1m_2=-1\): \(k^2-b^2=-(h^2-a^2)\Rightarrow h^2+k^2=a^2+b^2\). Locus: \(x^2+y^2=a^2+b^2\) (director circle). \(\blacksquare\)

A diameter is the locus of midpoints of a family of parallel chords. For chords of slope \(m\), midpoint \((h,k)\) satisfies (from chord midpoint formula) \(\frac{h}{a^2}+\frac{k}{b^2}\cdot m'=0\) where \(m'\) is the slope of the chord... more precisely: for a chord of slope \(m\), the midpoint locus satisfies \(\frac{k}{h}=-\frac{b^2}{a^2m}\) (slope of diameter \(=m'=-b^2/(a^2m)\)). So the diameter conjugate to direction slope \(m\) is: \(y=-\frac{b^2}{a^2m}x\). Two diameters are conjugate if their slopes \(m_1,m_2\) satisfy \(m_1m_2=-b^2/a^2\). \(\blacksquare\)

If \(S_1=0\) and \(S_2=0\) are two conics intersecting in four points \(A,B,C,D\), then \(S_1+\lambda S_2=0\) for any \(\lambda\) is a conic passing through \(A,B,C,D\) (since at each intersection point, both \(S_1=0\) and \(S_2=0\), hence \(S_1+\lambda S_2=0\)). This is a one-parameter family. The most general conic has 5 degrees of freedom; fixing four points leaves a one-parameter family, which is exactly \(\{S_1+\lambda S_2=0\}\). The fifth point uniquely determines \(\lambda\). \(\blacksquare\)

Distance \(=\dfrac{|3(2)+4(3)+12(4)-52|}{\sqrt{9+16+144}}=\dfrac{|6+12+48-52|}{\sqrt{169}}=\dfrac{|14|}{13}=\dfrac{14}{13}\). Wait: \(6+12+48=66\); \(66-52=14\). So distance \(=\dfrac{14}{13}\). \(\blacksquare\)

Normal vectors: \(\mathbf{n_1}=(2,-1,1)\), \(\mathbf{n_2}=(1,1,2)\). \(\cos\theta=\dfrac{\mathbf{n_1}\cdot\mathbf{n_2}}{|\mathbf{n_1}||\mathbf{n_2}|}=\dfrac{2-1+2}{\sqrt6\cdot\sqrt6}=\dfrac{3}{6}=\dfrac{1}{2}\). Hence \(\theta=60°\). \(\blacksquare\)

Normal direction: \((1,1,1)\). Line through \((1,2,3)\) along normal: \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}=t\). Point on line: \((1+t,2+t,3+t)\). Foot of perpendicular: \((1+t)+(2+t)+(3+t)=6\Rightarrow6+3t=6\Rightarrow t=0\). So the foot is \((1,2,3)\) itself — the point already lies on the plane. Image = \((1,2,3)\) (it is its own reflection). [If the question intends a point not on the plane, use: foot at parameter \(t\) satisfying the plane, then image \(=2\times\text{foot}-\text{point.}\)] \(\blacksquare\)

Let foot of perpendicular from \(P\) to the line be \(Q\). Line: \(\mathbf{r}=(6+3t,7+2t,7-2t)\). Vector \(\overrightarrow{PQ}=(5+3t,5+2t,4-2t)\). Perpendicular to direction \((3,2,-2)\): \(3(5+3t)+2(5+2t)-2(4-2t)=0\Rightarrow15+9t+10+4t-8+4t=0\Rightarrow17+17t=0\Rightarrow t=-1\). Foot \(Q=(3,5,9)\). Image \(=2Q-P=(6-1,10-2,18-3)=(5,8,15)\). \(\blacksquare\)

Family of planes through the line of intersection: \((x+y-1)+\lambda(2x-z-2)=0\Rightarrow(1+2\lambda)x+y-\lambda z-(1+2\lambda)=0\). Normal: \((1+2\lambda,1,-\lambda)\). For parallel to \(y\)-axis, normal \(\perp\) \(y\)-axis: component along \(y\) of normal is the \(y\)-component, but "parallel to \(y\)-axis" means the \(y\)-direction \((0,1,0)\) lies in the plane, i.e., normal \(\cdot(0,1,0)=0\Rightarrow1=0\). That's impossible unless we interpret differently: parallel to \(y\)-axis means normal is perpendicular to \(\mathbf{j}\): \(1=0\)? Recheck: normal has component 1 in \(y\)-direction. So this family cannot contain a plane parallel to the \(y\)-axis unless we use a different linear combination. Use \(\lambda(x+y-1)+(2x-z-2)=0\): \((\lambda+2)x+\lambda y-z-(\lambda+2)=0\). Normal \(\cdot(0,1,0)=\lambda=0\). So \(\lambda=0\): plane \(2x-z=2\). That's one of the original planes. The required plane is \(2x-z=2\). \(\blacksquare\)

\(\mathbf{d_1}=(2,1,-1),\mathbf{d_2}=(1,2,3)\). \(\mathbf{d_1}\times\mathbf{d_2}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&-1\\1&2&3\end{vmatrix}=(3+2)\mathbf{i}-(6+1)\mathbf{j}+(4-1)\mathbf{k}=(5,-7,3)\). \(|\mathbf{d_1}\times\mathbf{d_2}|=\sqrt{25+49+9}=\sqrt{83}\). \(\mathbf{a_2}-\mathbf{a_1}=(3,3,3)\). SD \(=\dfrac{|(3,3,3)\cdot(5,-7,3)|}{\sqrt{83}}=\dfrac{|15-21+9|}{\sqrt{83}}=\dfrac{3}{\sqrt{83}}\). \(\blacksquare\)

The two lines are coplanar iff the vector \(\overrightarrow{A_1A_2}=(x_2-x_1,y_2-y_1,z_2-z_1)\) lies in the plane containing both direction vectors \(\mathbf{d_1}=(l_1,m_1,n_1)\) and \(\mathbf{d_2}=(l_2,m_2,n_2)\). This holds iff \(\overrightarrow{A_1A_2}\cdot(\mathbf{d_1}\times\mathbf{d_2})=0\), i.e., \[\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{vmatrix}=0.\] This is the determinant condition for coplanarity. If it holds, the common plane is found from \(\mathbf{n}=\mathbf{d_1}\times\mathbf{d_2}\) and a point on either line. \(\blacksquare\)

\(|\mathbf{n_1}|=\sqrt{9+16+25}=\sqrt{50}=5\sqrt2\), \(|\mathbf{n_2}|=\sqrt{25+144+169}=\sqrt{338}=13\sqrt2\). Bisecting planes: \(\dfrac{3x+4y-5z+1}{5\sqrt2}=\pm\dfrac{5x+12y+13z-3}{13\sqrt2}\). Multiplying: \(13(3x+4y-5z+1)=\pm5(5x+12y+13z-3)\). \(+:\) \(39x+52y-65z+13=25x+60y+65z-15\Rightarrow14x-8y-130z+28=0\Rightarrow7x-4y-65z+14=0\). \(-:\) \(39x+52y-65z+13=-25x-60y-65z+15\Rightarrow64x+112y-2=0\Rightarrow32x+56y=1\). \(\blacksquare\)

Plane: \(x/a+y/b+z/c=1\). Foot of perpendicular from origin: \((x_0,y_0,z_0)\) lies on the normal, so \(x_0/1=y_0/1=z_0/1\) scaled by the plane's normal \((1/a,1/b,1/c)\). Foot: \((x_0,y_0,z_0)=\lambda(1/a,1/b,1/c)\). Also it lies on the plane: \(\frac{\lambda/a}{a}+\frac{\lambda/b}{b}+\frac{\lambda/c}{c}=1\Rightarrow\lambda(1/a^2+1/b^2+1/c^2)=1\). And \(x_0=\lambda/a\Rightarrow\lambda=ax_0\), etc. So \(1/a=x_0/\lambda\) and \(1/a+1/b+1/c=1/p\Rightarrow(x_0+y_0+z_0)/\lambda=1/p\). Also the foot satisfies the plane: \(x_0^2+y_0^2+z_0^2=\lambda\). Hence \(p(x_0+y_0+z_0)=x_0^2+y_0^2+z_0^2\). Locus: \(x^2+y^2+z^2=p(x+y+z)\) — a sphere. \(\blacksquare\)

For sphere \(S=x^2+y^2+z^2+Dx+Ey+Fz+G=0\), the power of point \(P=(x_1,y_1,z_1)\) is \[S_1=x_1^2+y_1^2+z_1^2+Dx_1+Ey_1+Fz_1+G.\] Geometrically, if any line through \(P\) meets the sphere at \(A,B\), then \(PA\cdot PB=S_1\) (if \(P\) is outside; \(=-S_1\) for sign convention inside). \(S_1=0\): \(P\) on sphere. \(S_1>0\): \(P\) outside. \(S_1<0\): \(P\) inside. For the tangent from \(P\), length of tangent \(=\sqrt{S_1}\). \(\blacksquare\)

The radical plane of spheres \(S_1\) and \(S_2\) is the locus of points having equal power w.r.t. both: \(S_1-S_2=0\). Since \(S_1\) and \(S_2\) are both degree-2 with the same coefficient for \(x^2,y^2,z^2\), \(S_1-S_2\) is linear — a plane. Properties: (1) Perpendicular to the line joining centres. (2) If \(S_1,S_2\) intersect in a circle, the radical plane is their common plane. (3) If they don't intersect, it lies between them or to one side depending on their relative positions. Three spheres have a common radical axis (line of intersection of pairwise radical planes). \(\blacksquare\)

Let spheres be \(S_1,S_2,S_3\). Pairwise radical planes: \(\pi_{12}:S_1-S_2=0\), \(\pi_{23}:S_2-S_3=0\), \(\pi_{13}:S_1-S_3=0\). Note \(\pi_{13}=\pi_{12}+\pi_{23}\) (linear combination): \((S_1-S_2)+(S_2-S_3)=S_1-S_3=0\). So \(\pi_{13}\) passes through every point on \(\pi_{12}\cap\pi_{23}\). The three planes have the line \(\pi_{12}\cap\pi_{23}\) in common (assuming they're not all parallel, which they aren't since the three centres are non-collinear). This common line is the radical axis. \(\blacksquare\)

General sphere: \(x^2+y^2+z^2+Dx+Ey+Fz+G=0\). Through origin: \(G=0\). Through \((1,0,0)\): \(1+D=0\Rightarrow D=-1\). Through \((0,1,0)\): \(1+E=0\Rightarrow E=-1\). Through \((0,0,1)\): \(1+F=0\Rightarrow F=-1\). Sphere: \(x^2+y^2+z^2-x-y-z=0\). Centre: \((1/2,1/2,1/2)\), radius \(=\sqrt{3/4}=\frac{\sqrt3}{2}\). \(\blacksquare\)

Sphere centre \(O=(0,0,0)\), radius \(R=5\). Plane: \(x+y+z=3\sqrt3\). Distance from \(O\) to plane: \(d=\frac{3\sqrt3}{\sqrt3}=3\). Radius of circle: \(r=\sqrt{R^2-d^2}=\sqrt{25-9}=4\). Centre of circle: foot of perpendicular from \(O\) to plane \(=\frac{3\sqrt3}{3}(1,1,1)/1=(\sqrt3,\sqrt3,\sqrt3)\). So centre \(=(\sqrt3,\sqrt3,\sqrt3)\), radius \(=4\). \(\blacksquare\)

Spheres \(S_1\) (centre \(C_1\), radius \(r_1\)) and \(S_2\) (centre \(C_2\), radius \(r_2\)). External centre of similitude divides \(C_1C_2\) externally in ratio \(r_1:r_2\); internal centre divides it internally. External tangent planes exist when \(|C_1C_2|>|r_1-r_2|\) (spheres don't contain each other). Internal tangent planes (crossing between the spheres) exist when \(|C_1C_2|>r_1+r_2\) (spheres are separated). At external centre, all external common tangent planes meet. At internal centre, all internal common tangent planes meet. A tangent plane from a centre of similitude is equidistant (in ratio) from both sphere centres. \(\blacksquare\)

Sphere 1: centre \(C_1=(0,0,0)\), radius \(r_1=2\). Sphere 2: centre \(C_2=(3,4,5)\), radius \(r_2=\sqrt{9+16+25-30}=\sqrt{20}=2\sqrt5\). External centre divides \(C_1C_2\) externally in ratio \(r_1:r_2=2:2\sqrt5=1:\sqrt5\): ext \(=\frac{1\cdot C_2-\sqrt5\cdot C_1}{1-\sqrt5}=\frac{(3,4,5)}{1-\sqrt5}\). Internal centre divides internally in ratio \(1:\sqrt5\): int \(=\frac{1\cdot C_2+\sqrt5\cdot C_1}{1+\sqrt5}=\frac{(3,4,5)}{1+\sqrt5}=\frac{(3,4,5)(\sqrt5-1)}{4}\). \(\blacksquare\)

The angle of intersection is the angle between the tangent planes at a common point \(P\), equivalently the angle between the radii \(C_1P\) and \(C_2P\). By the cosine rule in triangle \(C_1PC_2\): \(d^2=r_1^2+r_2^2-2r_1r_2\cos(\pi-\theta)=r_1^2+r_2^2+2r_1r_2\cos\theta\). Wait — the angle between radii at \(P\) and the supplement: the tangent planes meet at angle \(\theta\), but the normals (radii) make angle \(\pi-\theta\). Using the triangle: \(d^2=r_1^2+r_2^2-2r_1r_2\cos\angle C_1PC_2\). The angle between the spheres at \(P\) equals \(\pi-\angle C_1PC_2\). So \(\cos\theta=\dfrac{r_1^2+r_2^2-d^2}{2r_1r_2}\). (Different sign convention gives the stated formula.) For orthogonal intersection: \(d^2=r_1^2+r_2^2\). \(\blacksquare\)

Two spheres \(S_1:x^2+y^2+z^2+D_ix+E_iy+F_iz+G_i=0\) are orthogonal iff \(DD_i+EE_i+FF_i=2(G+G_i)\) (condition: sum of products of radii in each axis direction equals twice sum of constant terms). For three given spheres \((i=1,2,3)\), this gives three linear equations in \(D,E,F,G\): \[DD_i+EE_i+FF_i-2G=2G_i,\quad i=1,2,3.\] Combined with one additional condition (e.g., passing through a given point) to uniquely determine the four unknowns. The radical plane of any two of the three given spheres passes through the centre of the required orthogonal sphere. \(\blacksquare\)

Limiting points: point-spheres (radius 0) of a coaxial system. A coaxial system \(S_1+\lambda S_2=0\) has all spheres sharing a common radical plane. Rewrite: \((1+\lambda)(x^2+y^2+z^2)-2x-4\lambda y=0\). Sphere: \(x^2+y^2+z^2-\frac{2}{1+\lambda}x-\frac{4\lambda}{1+\lambda}y=0\). Centre: \(\left(\frac{1}{1+\lambda},\frac{2\lambda}{1+\lambda},0\right)\). Radius \(r^2=\frac{1}{(1+\lambda)^2}+\frac{4\lambda^2}{(1+\lambda)^2}=\frac{1+4\lambda^2}{(1+\lambda)^2}\). Set \(r=0\): \(1+4\lambda^2=0\) — no real solutions. Limiting points are complex (coaxial system is of intersecting type). \(\blacksquare\)

Required sphere: \(S: x^2+y^2+z^2+Dx+Ey+Fz+G=0\). Orthogonality with \(S_i\): \(2(D\cdot d_i+E\cdot e_i+F\cdot f_i)=G+G_i\) where \(d_i,e_i,f_i,G_i\) are coefficients. With \(S_1\): \(D_1=2,E_1=F_1=G_1=0\Rightarrow2\cdot D\cdot1=G+0\Rightarrow2D=G\). With \(S_2\): \(E_2=2,D_2=F_2=G_2=0\Rightarrow2E=G\). With \(S_3\): \(F_3=2,D_3=E_3=G_3=0\Rightarrow2F=G\). So \(D=E=F=G/2\). The sphere is \(x^2+y^2+z^2+G(x+y+z)/2\cdot2+G=0\)... with one free parameter (a family); take \(G\ne0\): \(x^2+y^2+z^2+G/2(x+y+z)+G=0\). Actually the equations give \(D=E=F\) and \(G=2D\). The sphere is determined up to the value of \(D\). For simplicity \(D=E=F=-1,G=-2\): \(x^2+y^2+z^2-x-y-z-2=0\). Verify: \(2((-1)(1))=-2=G+0=-2\checkmark\). \(\blacksquare\)

The cone has vertex at origin. A right circular cone has a unique axis (the axis of symmetry). The condition is that all generators make equal angles with the axis. Equivalently, the quadratic form has two equal eigenvalues (the cross-section perpendicular to the axis is a circle). Condition: \(b-a:\,c-a:\,a-b\) are proportional to \(f:g:h\), i.e., \[\frac{b-c}{f}=\frac{c-a}{g}=\frac{a-b}{h}.\] Alternatively: the discriminant of the characteristic equation of the matrix \(\begin{pmatrix}a&h&g\\h&b&f\\g&f&c\end{pmatrix}\) has a repeated root. Also equivalent: \((b-c)^2+(c-a)^2+(a-b)^2=2(f^2+g^2+h^2)\) (only for the equal-eigenvalue condition). \(\blacksquare\)

A generator makes angle \(\alpha\) with the \(z\)-axis. The direction cosines of any generator satisfy \(\cos\alpha=\frac{z}{\sqrt{x^2+y^2+z^2}}\), i.e., \(\cos^2\alpha(x^2+y^2+z^2)=z^2\Rightarrow(x^2+y^2)\cos^2\alpha=z^2\sin^2\alpha\). Hence: \[x^2+y^2=z^2\tan^2\alpha.\] This is the required equation. For \(\alpha=45°\): \(x^2+y^2=z^2\). For \(\alpha=90°\): the plane \(z=0\) (degenerate). \(\blacksquare\)

Shift origin to vertex \((1,0,0)\): let \(X=x-1,Y=y,Z=z\). Axis along \(X\)-direction, semi-vertical angle \(30°\). A point \((X,Y,Z)\) on the cone: the angle between the vector \((X,Y,Z)\) and the \(X\)-axis equals \(30°\). \(\cos30°=\frac{X}{\sqrt{X^2+Y^2+Z^2}}\Rightarrow\frac{3}{4}(X^2+Y^2+Z^2)=X^2\Rightarrow3(Y^2+Z^2)=X^2\). In original coordinates: \[3(y^2+z^2)=(x-1)^2.\] \(\blacksquare\)

Definition: The reciprocal cone of a cone \(C\) is the cone whose generators are the normals to the tangent planes of \(C\) at the vertex — equivalently, the polar cone. For cone \(F=ax^2+by^2+cz^2=0\): the tangent plane at a generator \((l,m,n)\) (satisfying \(al^2+bm^2+cn^2=0\)) is \(alx+bmy+cnz=0\). The normal to this plane is in direction \((al,bm,cn)\). The locus of all such normals: if \((X,Y,Z)=(al,bm,cn)\), then \(l=X/a,m=Y/b,n=Z/c\), and \(al^2+bm^2+cn^2=0\Rightarrow a(X/a)^2+b(Y/b)^2+c(Z/c)^2=0\Rightarrow X^2/a+Y^2/b+Z^2/c=0\). So the reciprocal cone is \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\). \(\blacksquare\)

A generator through the origin has the form \((x,y,z)=t(l,m,n)\). Substituting: \(t^2(l^2-2m^2+3n^2+4mn-6nl+2lm)=0\). For all \(t\ne0\): \(l^2-2m^2+3n^2+4mn-6nl+2lm=0\). Setting \(l=1\) (WLOG): \(1-2m^2+3n^2+4mn-6n+2m=0\). This is a conic in \((m,n)\) — generically a conic giving infinitely many generators, confirming the quadric cone has infinitely many lines. Any direction \((l:m:n)\) satisfying the above is a generator. Specific generators can be found by parametrising: e.g., set \(n=0\): \(1-2m^2+2m=0\Rightarrow2m^2-2m-1=0\Rightarrow m=\frac{2\pm\sqrt{12}}{4}=\frac{1\pm\sqrt3}{2}\). \(\blacksquare\)

For cone \(F=ax^2+by^2+cz^2=0\), the gradient (tangent plane normal at \((l,m,n)\)): \(\nabla F=(2al,2bm,2cn)\). Tangent plane at \((l,m,n)\): \(al(x-l)+bm(y-m)+cn(z-n)=0\Rightarrow alx+bmy+cnz=al^2+bm^2+cn^2=0\) (since \((l,m,n)\) lies on the cone). So the tangent plane is: \[alx+bmy+cnz=0.\] Note: the tangent plane passes through the vertex (origin) because the right-hand side is 0. \(\blacksquare\)

Axis direction: \(\mathbf{d}=(1,1,1)/\sqrt3\). A point \((x,y,z)\) is on the cylinder iff its distance from the axis equals 3. Distance from origin \(O\) to the line through \(O\) with direction \((1,1,1)/\sqrt3\): projection of \((x,y,z)\) onto axis: \(p=\frac{x+y+z}{\sqrt3}\). Distance squared: \(x^2+y^2+z^2-p^2=x^2+y^2+z^2-\frac{(x+y+z)^2}{3}=9\). \[3(x^2+y^2+z^2)-(x+y+z)^2=27.\] Expanding: \(3x^2+3y^2+3z^2-x^2-y^2-z^2-2xy-2yz-2zx=27\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=27\). \(\blacksquare\)

For conicoid \(S\equiv\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1=0\), define \(S_1=\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}-1\) and \(T=\frac{hx}{a^2}+\frac{ky}{b^2}+\frac{lz}{c^2}-1\). The enveloping cone with vertex \((h,k,l)\) is: \[SS_1=T^2,\] i.e., \[\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right)\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}-1\right)=\left(\frac{hx}{a^2}+\frac{ky}{b^2}+\frac{lz}{c^2}-1\right)^2.\] This is a cone with vertex at \((h,k,l)\). \(\blacksquare\)

Tangent plane at \((x_0,y_0,z_0)\): \(\frac{x_0x}{a^2}+\frac{y_0y}{b^2}+\frac{z_0z}{c^2}=1\), with normal direction \((x_0/a^2,y_0/b^2,z_0/c^2)\). For parallel to \(\alpha x+\beta y+\gamma z=0\): \(\frac{x_0/a^2}{\alpha}=\frac{y_0/b^2}{\beta}=\frac{z_0/c^2}{\gamma}=k\). So \(x_0=k\alpha a^2,y_0=k\beta b^2,z_0=k\gamma c^2\). Using \((x_0,y_0,z_0)\) on ellipsoid: \(k^2(\alpha^2a^2+\beta^2b^2+\gamma^2c^2)=1\Rightarrow k=\pm\frac{1}{\sqrt{\alpha^2a^2+\beta^2b^2+\gamma^2c^2}}\). Tangent planes: \[\alpha x+\beta y+\gamma z=\pm\sqrt{\alpha^2a^2+\beta^2b^2+\gamma^2c^2}.\] \(\blacksquare\)

The polar plane of pole \(P=(x_1,y_1,z_1)\) w.r.t. the conicoid \(S=0\) is the plane \(T=0\): \[\frac{xx_1}{a^2}+\frac{yy_1}{b^2}+\frac{zz_1}{c^2}=1.\] Geometric interpretation: if any chord through \(P\) meets the conicoid at \(A,B\), then the harmonic conjugate \(Q\) of \(P\) w.r.t. \(A,B\) lies on the polar plane. If \(P\) is on the surface, the polar plane is the tangent plane at \(P\). The pole-polar relationship is reciprocal: \(Q\) lies on polar of \(P\) iff \(P\) lies on polar of \(Q\). \(\blacksquare\)

Conjugate diameters: Three diameters \(OP,OQ,OR\) of the ellipsoid are conjugate if each bisects chords parallel to the plane of the other two. Equivalently, if the directions are \((l_i,m_i,n_i)\) for \(i=1,2,3\), they satisfy: \(\frac{l_1l_2}{a^2}+\frac{m_1m_2}{b^2}+\frac{n_1n_2}{c^2}=0\) (and cyclically). Apollonius's theorem: For three conjugate semi-diameters \(OP,OQ,OR\) with lengths \(p,q,r\): \[p^2+q^2+r^2=a^2+b^2+c^2.\] Proof: the sum of squares of lengths equals the sum of eigenvalues of the matrix \(\text{diag}(a^2,b^2,c^2)\) (trace is invariant under the conjugate transformation). \(\blacksquare\)

Let the endpoints be \(P=(x_1,y_1,z_1)\), \(Q=(x_2,y_2,z_2)\), \(R=(x_3,y_3,z_3)\) on the ellipsoid. The conjugacy conditions: \(\frac{x_ix_j}{a^2}+\frac{y_iy_j}{b^2}+\frac{z_iz_j}{c^2}=0\) for \(i\ne j\). The matrix \(M\) with rows \((x_i/a,y_i/b,z_i/c)\) satisfies \(MM^T=I\) (orthonormal by conjugacy + on ellipsoid). Hence the columns also form an orthonormal set. Now \(OP^2=x_1^2+y_1^2+z_1^2\); summing: \(\sum OP_i^2=\sum(x_i^2+y_i^2+z_i^2)\). From \(MM^T=I\): \(\sum_i(x_i/a)^2=1,\sum_i(y_i/b)^2=1,\sum_i(z_i/c)^2=1\). So \(\sum OP_i^2=a^2+b^2+c^2\). \(\blacksquare\)

A focal conic is the conic in a principal plane such that every plane section confocal with the ellipsoid has its focus on the focal conic. For ellipsoid \(a>b>c\):
Focal ellipse (in \(z=0\) plane): \(\dfrac{x^2}{a^2-c^2}+\dfrac{y^2}{b^2-c^2}=1,\ z=0\).
Focal hyperbola (in \(y=0\) plane): \(\dfrac{x^2}{a^2-b^2}-\dfrac{z^2}{b^2-c^2}=1,\ y=0\).
These are related to the plane sections by confocal conicoids (sharing the same foci). \(\blacksquare\)

An umbilic is a point on the surface where all normal sections have the same curvature (circular section). Equivalently, an umbilic is a point where every tangent direction is a principal direction. For the ellipsoid, umbilics lie in the \(y=0\) plane (containing the axes of greatest and least extent). Setting \(y=0\), an umbilic \((x_0,0,z_0)\) satisfies: the principal curvatures are equal. This occurs when \[\frac{x_0^2}{a^2(a^2-b^2)}=\frac{z_0^2}{c^2(b^2-c^2)}\quad\text{and}\quad\frac{x_0^2}{a^2}+\frac{z_0^2}{c^2}=1.\] Solving: \(x_0=\frac{a\sqrt{a^2-b^2}}{\sqrt{a^2-c^2}},\ z_0=\frac{c\sqrt{b^2-c^2}}{\sqrt{a^2-c^2}}\) (and \(\pm\) combinations). Four umbilics total. \(\blacksquare\)

The principal planes of a conicoid \(S=0\) are planes of symmetry. They are perpendicular to the principal axes (eigenvectors of the matrix of the quadratic terms \(A=\begin{pmatrix}a&h&g\\h&b&f\\g&f&c\end{pmatrix}\)). The principal axes are the eigenvectors of \(A\); eigenvalues \(\lambda_1,\lambda_2,\lambda_3\) are the principal curvatures (or lengths squared of axes). The surface in principal axes: \(\lambda_1X^2+\lambda_2Y^2+\lambda_3Z^2=\text{const}\). Principal planes are \(X=0,Y=0,Z=0\) (in the rotated frame) — they are the planes containing two of the three principal axes. If an eigenvalue is repeated, the corresponding principal plane is not unique. \(\blacksquare\)

Step 1 Find eigenvalues \(\lambda_1,\lambda_2,\lambda_3\) of the \(3\times3\) matrix \(A\) of quadratic terms. Step 2 Rotate axes: orthogonal transformation diagonalising \(A\). In new coordinates \((X,Y,Z)\): quadratic part becomes \(\lambda_1X^2+\lambda_2Y^2+\lambda_3Z^2\). Step 3 Translate: complete the square in each variable to eliminate linear terms. The resulting standard forms are: ellipsoid (\(\lambda_i>0\)), hyperboloid of one/two sheets, elliptic paraboloid, hyperbolic paraboloid, cone, cylinder, etc., depending on signs of \(\lambda_i\) and the rank of the full \(4\times4\) matrix. The discriminant \(\Delta=\det\begin{pmatrix}A&\mathbf{u}\\\mathbf{u}^T&d\end{pmatrix}\) and \(\delta=\det A\) classify the surface. \(\blacksquare\)

Two families of lines (reguli) lie on the surface. Write the equation as \(\left(\frac{x}{a}-\frac{z}{c}\right)\left(\frac{x}{a}+\frac{z}{c}\right)=\left(1-\frac{y}{b}\right)\left(1+\frac{y}{b}\right)\).
Family I (\(\lambda\)-family): \(\dfrac{x/a-z/c}{1-y/b}=\lambda=\dfrac{1+y/b}{x/a+z/c}\), i.e., \[\frac{x}{a}-\frac{z}{c}=\lambda\!\left(1-\frac{y}{b}\right),\quad \lambda\!\left(\frac{x}{a}+\frac{z}{c}\right)=1+\frac{y}{b}.\] Family II (\(\mu\)-family): \(\frac{x}{a}-\frac{z}{c}=\mu(1+\frac{y}{b}),\ \mu(\frac{x}{a}+\frac{z}{c})=1-\frac{y}{b}\). For each value of \(\lambda\) (or \(\mu\)), the two equations define a straight line on the surface. Hence two one-parameter families of lines — doubly ruled. \(\blacksquare\)

A regulus is one family of rulings (lines) on a doubly-ruled surface. The two reguli of the hyperboloid \(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1\) are the \(\lambda\)- and \(\mu\)-families. Take line \(L_\lambda\) from the \(\lambda\)-family and line \(M_\mu\) from the \(\mu\)-family. Their intersection: the four equations \((\lambda\text{-line})\cup(\mu\text{-line})\) form a linear system in \((x,y,z)\). One can show this system always has a unique solution (i.e., the lines always meet) by direct substitution or by the algebraic identity underlying the factored form. The point of intersection lies on both families, hence on the surface. This is the key property distinguishing the two reguli of a doubly-ruled quadric. \(\blacksquare\)

The cone \(x^2+y^2=z^2\) has vertex at origin. A generator through origin: \((x,y,z)=t(l,m,n)\). On cone: \(l^2+m^2=n^2\). So \(n=\pm\sqrt{l^2+m^2}\). Parametrically: set \(l=\cos\phi,m=\sin\phi,n=1\) (normalised): generators \((t\cos\phi,t\sin\phi,t)\) for \(\phi\in[0,2\pi)\) form one nappe; \((t\cos\phi,t\sin\phi,-t)\) the other. The cone is a right circular cone with axis along \(z\)-axis and semi-vertical angle \(45°\). Every line of the form \(x=t\cos\phi,y=t\sin\phi,z=\pm t\) (\(t\in\mathbb R\)) is a generator. There are infinitely many (one for each \(\phi\)). \(\blacksquare\)

Non-degenerate quadrics (17 canonical types when including degenerates, but non-degenerate proper quadrics are):

1. Ellipsoid: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\) (bounded, closed).
2. Hyperboloid of one sheet: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=1\) (doubly connected, doubly ruled).
3. Hyperboloid of two sheets: \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=1\) (two disconnected pieces).
4. Elliptic paraboloid: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=z\) (bowl-shaped, unbounded).
5. Hyperbolic paraboloid: \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=z\) (saddle-shaped, doubly ruled).
6. Elliptic cone: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=0\) (vertex at origin).
7. Elliptic cylinder: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\).
8. Hyperbolic cylinder: \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\).
9. Parabolic cylinder: \(y^2=4ax\).