UPSC 2020 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

Ellipsoid · Analytic Geometry · asked 3× in 13 yrs · Read the full method →

Question

Find the equations of the tangent plane to the ellipsoid $2x^2+6y^2+3z^2=27$ which passes through the line $x-y-z=0=x-y+2z-9$ .

Pencil of planes through the line of intersection; impose the ellipsoid tangency condition $a^2l^2+b^2m^2+c^2n^2=p^2$ .

The line is the intersection of $x-y-z=0$ and $x-y+2z-9=0$ . Any plane through it is

(x-y-z)+\lambda(x-y+2z-9)=0,

i.e.

(1+\lambda)x-(1+\lambda)y+(2\lambda-1)z=9\lambda.

Write this as $lx+my+nz=p$ with

l=1+\lambda,\quad m=-(1+\lambda),\quad n=2\lambda-1,\quad p=9\lambda.

Put the ellipsoid in standard form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ :

\frac{x^2}{27/2}+\frac{y^2}{27/6}+\frac{z^2}{27/3}=1\ \Rightarrow\ a^2=\tfrac{27}{2},\ b^2=\tfrac92,\ c^2=9.

A plane $lx+my+nz=p$ is tangent to this ellipsoid iff

a^2 l^2+b^2 m^2+c^2 n^2=p^2.

\frac{27}{2}(1+\lambda)^2+\frac{9}{2}(1+\lambda)^2+9(2\lambda-1)^2=(9\lambda)^2.

The first two terms combine: $\big(\tfrac{27}{2}+\tfrac92\big)(1+\lambda)^2=18(1+\lambda)^2$ . So

18(1+\lambda)^2+9(2\lambda-1)^2=81\lambda^2.

Divide by $9$ :

2(1+\lambda)^2+(2\lambda-1)^2=9\lambda^2.

Expand: $2(1+2\lambda+\lambda^2)+(4\lambda^2-4\lambda+1)=9\lambda^2$ , i.e. $6\lambda^2+3=9\lambda^2$ , so $3\lambda^2=3$ and

\lambda=\pm1.

$\lambda=1$ : $l=2,\ m=-2,\ n=1,\ p=9$ :

2x-2y+z=9.

$\lambda=-1$ : $l=0,\ m=0,\ n=-3,\ p=-9$ , i.e. $-3z=-9$ :

z=3.

\boxed{\;2x-2y+z=9\quad\text{and}\quad z=3.\;}