← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q2c-ii — Step-by-Step Solution

10 marks · Section A

Cone · Analytic Geometry · asked 14× in 13 yrs · Read the full method →

Question

Prove that the plane $z=0$ cuts the enveloping cone of the sphere $x^2+y^2+z^2=11$ which has the vertex at $(2,4,1)$ in a rectangular hyperbola.

Technique

Enveloping cone $SS_1=T^2$; set $z=0$; apply the rectangular-hyperbola criterion (coeff $x^2$ + coeff $y^2$ $=0$).

Solution

Step 1 — Enveloping cone $SS_1=T^2$

For the sphere $S\equiv x^2+y^2+z^2-11=0$ and vertex $P=(x_1,y_1,z_1)=(2,4,1)$, the enveloping cone (the locus of tangent lines from $P$) is

where

Step 2 — Write out the cone

Expanding the right side:

So the cone is

i.e.

Step 3 — Intersect with $z=0$

Set $z=0$:

Answer