UPSC 2021 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

Cylinder · Analytic Geometry · asked 4× in 13 yrs · Read the full method →

Question

Find the equation of the cylinder whose generators are parallel to the line $x=-\dfrac{y}{2}=\dfrac{z}{3}$ and whose guiding curve is $x^2+2y^2=1$ , $z=0$ .

Technique

Parametrise points on the cylinder via guiding curve point + multiple of generator direction; eliminate parameter using the curve equation.

Solution

Setup. Generators parallel to direction $(1,-2,3)$ (from $x/1=y/(-2)=z/3$ ).

Strategy. A point $(X,Y,Z)$ on the cylinder lies on some generator that passes through a point $(x_0,y_0,0)$ on the guiding curve (the ellipse $x^2+2y^2=1$ in the $z=0$ plane). The generator’s direction is $(1,-2,3)$ .

Step 1 — Parametrize the generator

Generator through $(x_0,y_0,0)$ with direction $(1,-2,3)$ : $(X,Y,Z)=(x_0+t,\;y_0-2t,\;3t)$ .

From the $Z$ -coordinate: $t=Z/3$ .

So $x_0=X-Z/3$ , $y_0=Y+2Z/3$ .

Step 2 — Use guiding curve condition

$(x_0,y_0)$ is on $x_0^2+2y_0^2=1$ :

$(X-Z/3)^2+2(Y+2Z/3)^2=1$ .

Step 3 — Expand

$(X-Z/3)^2=X^2-\dfrac{2XZ}{3}+\dfrac{Z^2}{9}$ .

$2(Y+2Z/3)^2=2[Y^2+\dfrac{4YZ}{3}+\dfrac{4Z^2}{9}]=2Y^2+\dfrac{8YZ}{3}+\dfrac{8Z^2}{9}$ .

Sum: $X^2-\dfrac{2XZ}{3}+\dfrac{Z^2}{9}+2Y^2+\dfrac{8YZ}{3}+\dfrac{8Z^2}{9}=X^2+2Y^2+\dfrac{9Z^2}{9}-\dfrac{2XZ}{3}+\dfrac{8YZ}{3}$

$=X^2+2Y^2+Z^2-\dfrac{2XZ}{3}+\dfrac{8YZ}{3}=1$ .

Multiply by 3: $3X^2+6Y^2+3Z^2-2XZ+8YZ=3$ .

(Replace $X,Y,Z$ with $x,y,z$ for the cylinder equation.)

Answer

\boxed{\;3x^2+6y^2+3z^2-2xz+8yz=3.\;}