UPSC 2020 Maths Optional Paper 1 Q4b — Step-by-Step Solution

15 marks · Section A

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

Question

Find the locus of the point of intersection of the perpendicular generators of the hyperbolic paraboloid $\frac{x^2}{a^2}-\frac{y^2}{b^2}=2z$ .

Technique

Parametrize the two rulings, get direction vectors via cross products of plane normals, impose orthogonality, and evaluate $z$ at the meeting point.

Solution

Step 1 — The two families of generators

Factor the surface equation:

\left(\frac{x}{a}-\frac{y}{b}\right)\left(\frac{x}{a}+\frac{y}{b}\right)=2z.

$\lambda$ -family (parameter $\lambda$ ):

\frac{x}{a}-\frac{y}{b}=2\lambda,\qquad \lambda\left(\frac{x}{a}+\frac{y}{b}\right)=z.

$\mu$ -family (parameter $\mu$ ):

\frac{x}{a}+\frac{y}{b}=2\mu,\qquad \mu\left(\frac{x}{a}-\frac{y}{b}\right)=z.

(Each value of the parameter, multiplied out, returns the surface equation, so these are rulings.)

Step 2 — Direction vectors

Each generator is the intersection of two planes; its direction is the cross product of the normals.

$\lambda$ -generator: normals $\left(\tfrac1a,-\tfrac1b,0\right)$ and $\left(\tfrac{\lambda}{a},\tfrac{\lambda}{b},-1\right)$ . Cross product (times $ab$ ):

\mathbf{d}_1=(a,\ b,\ 2\lambda).

$\mu$ -generator: normals $\left(\tfrac1a,\tfrac1b,0\right)$ and $\left(\tfrac{\mu}{a},-\tfrac{\mu}{b},-1\right)$ . Cross product (times $ab$ ):

\mathbf{d}_2=(-a,\ b,\ -2\mu).

Step 3 — Perpendicularity condition

\mathbf{d}_1\cdot\mathbf{d}_2=(a)(-a)+(b)(b)+(2\lambda)(-2\mu)=-a^2+b^2-4\lambda\mu=0,

\Rightarrow\ 4\lambda\mu=b^2-a^2.\tag{$\ast$}

Step 4 — Coordinates of the intersection point

At a point $P=(x,y,z)$ lying on a $\lambda$ -generator and a $\mu$ -generator,

\frac{x}{a}-\frac{y}{b}=2\lambda,\qquad \frac{x}{a}+\frac{y}{b}=2\mu.

Its $z$ -coordinate (from the $\lambda$ -family’s second equation):

z=\lambda\left(\frac{x}{a}+\frac{y}{b}\right)=\lambda\cdot2\mu=2\lambda\mu.

Using $(\ast)$ , $2\lambda\mu=\dfrac{b^2-a^2}{2}$ , hence

z=\frac{b^2-a^2}{2}.

Step 5 — The locus

The $z$ -coordinate is constant for every such intersection point, with $x,y$ free; the locus is the plane

Answer

\boxed{\;2z=b^2-a^2\quad\Longleftrightarrow\quad z=\frac{b^2-a^2}{2}.\;}