← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

Family of surfaces · PDEs · asked 7× in 13 yrs · Read the full method →

Question

Find the partial differential equation of the family of all tangent planes to the ellipsoid $x^2+4y^2+4z^2=4$, which are not perpendicular to the $xy$-plane.

Technique

Tangency condition $A^2u^2+B^2v^2+C^2w^2=k^2$ for a plane and a central conicoid; write the plane as $z=px+qy+r$, eliminate $r=z-px-qy$ to get a Clairaut-type PDE.

Solution

Setup. Write the ellipsoid in standard form by dividing by $4$:

We seek the PDE whose complete integral is the two-parameter family of tangent planes. A plane that is not perpendicular to the $xy$-plane has a non-vertical normal, hence can be solved for $z$:

(A plane perpendicular to the $xy$-plane would contain the $z$-direction and could not be written this way — these are precisely the cases the question excludes.)

Step 1 — Tangency condition

For the plane $ux+vy+wz=k$ to touch the ellipsoid $\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}+\dfrac{z^2}{C^2}=1$, the condition is

Write the plane $z=px+qy+r$ as $px+qy-z+r=0$, i.e. $u=p,\ v=q,\ w=-1,\ k=-r$. Then

Step 2 — Eliminate the parameters $r$

Since $z=px+qy+r$, we have $r=z-px-qy$. Substituting into $r^2=4p^2+q^2+1$ gives the required PDE:

Answer