← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q4c — Step-by-Step Solution

13 marks · Section A

Paraboloid (elliptic and hyperbolic) · Analytic Geometry · asked 6× in 13 yrs · Read the full method →

Question

Two perpendicular tangent planes to the paraboloid $x^2+y^2=2z$ intersect in a straight line in the plane $x=0$. Obtain the curve to which this straight line touches.

Technique

Parametrise tangent planes to the paraboloid via point of tangency; impose perpendicularity of normals; impose “line lies in $x=0$”; reduce to a 1-parameter family of lines in the $yz$-plane; find envelope by $F=0$, $\partial F/\partial\beta=0$.

Solution

Setup. Paraboloid $S:\;x^2+y^2-2z=0$. A tangent plane to $S$ at $(x_0,y_0,z_0)\in S$ has equation

(Using $\nabla(x^2+y^2-2z)=(2x_0,2y_0,-2)$ as normal direction; the tangent-plane equation can also be derived via "$\dfrac{x x_0+y y_0}{2}=\dfrac{z+z_0}{2}$" — the chord-bisection rule for a quadric — giving $x_0 x+y_0 y=z+z_0$ as above.)

So a tangent plane has the form

where $(\alpha,\beta,c)$ corresponds to point of tangency $(\alpha,\beta,c)$ on the paraboloid (since $z_0=(\alpha^2+\beta^2)/2$).

Equivalently: $\pi$ is tangent to $S$ iff it can be written $\alpha x+\beta y-z=(\alpha^2+\beta^2)/2$ for some $\alpha,\beta$.

Step 1 — Two perpendicular tangent planes

Let $\pi_1:\;\alpha_1 x+\beta_1 y-z=(\alpha_1^2+\beta_1^2)/2$ and $\pi_2:\;\alpha_2 x+\beta_2 y-z=(\alpha_2^2+\beta_2^2)/2$.

Normals: $\vec n_1=(\alpha_1,\beta_1,-1)$, $\vec n_2=(\alpha_2,\beta_2,-1)$.

Perpendicularity: $\vec n_1\cdot\vec n_2=\alpha_1\alpha_2+\beta_1\beta_2+1=0$, i.e.

Step 2 — Line of intersection lies in $x=0$

The line $\pi_1\cap\pi_2$ lies entirely in the plane $x=0$ iff the $x$-coordinates of all points satisfying both equations are zero. Equivalently: setting $x=0$ in both equations gives a 1-parameter family in $(y,z)$, and this must coincide with $\pi_1\cap\pi_2$.

Necessary condition: $\pi_1$ and $\pi_2$ when restricted to $x=0$ give the same line in the $yz$-plane (otherwise their intersection in $x=0$ is a single point, not a line).

On $x=0$: $\pi_i$ reduces to $\beta_i y-z=(\alpha_i^2+\beta_i^2)/2$.

For these to be the same line: $\beta_1=\beta_2$ and $(\alpha_1^2+\beta_1^2)/2=(\alpha_2^2+\beta_2^2)/2$.

Let $\beta_1=\beta_2=\beta$. Then $\alpha_1^2=\alpha_2^2$, so $\alpha_2=\pm\alpha_1$. If $\alpha_2=\alpha_1$, the planes coincide — degenerate. So $\alpha_2=-\alpha_1$, giving distinct planes that meet on $x=0$.

Let $\alpha_1=\alpha$, $\alpha_2=-\alpha$, $\beta_1=\beta_2=\beta$.

Step 3 — Apply perpendicularity $(\#)$

$\alpha_1\alpha_2+\beta_1\beta_2=-\alpha^2+\beta^2=-1$, so

Step 4 — Equation of the line in $x=0$

Restricting to $x=0$, $\pi_1$ and $\pi_2$ both give $\beta y-z=(\alpha^2+\beta^2)/2$. So the line is

Using $(\diamond)$: $\alpha^2=1+\beta^2$, so $\alpha^2+\beta^2=1+2\beta^2$, and

Step 5 — Envelope (in the $yz$-plane $x=0$) as $\beta$ varies

The family of lines in the $yz$-plane is

Envelope: solve $F=0$ and $\partial F/\partial\beta=0$ simultaneously.

$\partial F/\partial\beta=-y+2\beta=0\Rightarrow\beta=y/2$.

Substitute: $z-(y/2)y+(y/2)^2+1/2=0\Rightarrow z-y^2/2+y^2/4+1/2=0\Rightarrow z=y^2/4-1/2$.

So the envelope is

Answer