UPSC Maths 2025 Paper 1 Q2c-i — Solution

10 marks · Section A

Question

Find the equation of the cylinder whose generators are parallel to the line $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$ and that passes through the curve $x^2 + y^2 = 16$ , $z = 0$ .

Technique

Treat the cylinder as the locus of lines (generators) of fixed direction through a guiding curve: project a general point back along the generator direction onto the plane of the guiding curve and substitute.

Solution

Step 1 — Generator through a general point.

The generators have direction $\vec d=(1,2,3)$ . Let $P=(X,Y,Z)$ be a general point of the cylinder. The generator through $P$ is $(x,y,z)=(X,Y,Z)+\lambda(1,2,3).$

Step 2 — Meet the plane $z=0$ of the guiding curve.

Set $z=0$ : $Z+3\lambda=0\Rightarrow \lambda=-\dfrac{Z}{3}$ . The intersection point has coordinates $x^*=X+\lambda = X-\frac{Z}{3},\qquad y^*=Y+2\lambda = Y-\frac{2Z}{3}.$

Step 3 — Impose the guiding curve $x^{*2}+y^{*2}=16$ .

$\left(X-\frac{Z}{3}\right)^2+\left(Y-\frac{2Z}{3}\right)^2=16.$ Multiply by $9$ : $(3X-Z)^2+(3Y-2Z)^2=144.$

Step 4 — Expand (rename $X,Y,Z\to x,y,z$ ): $9x^2-6xz+z^2+9y^2-12yz+4z^2=144$ $9x^2+9y^2+5z^2-6xz-12yz-144=0.$

Answer

$\boxed{\;9x^2 + 9y^2 + 5z^2 - 6xz - 12yz - 144 = 0.\;}$