← 2018 Paper 1

UPSC 2018 Maths Optional Paper 1 Q3c — Step-by-Step Solution

13 marks · Section A

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

Question

Find the equations to the generating lines of the paraboloid $(x+y+z)(2x+y-z)=6z$ which pass through the point $(1,1,1)$.

Technique

Ruled quadric: a line through a surface point lies on the surface iff the $t^1$ and $t^2$ coefficients of the substituted polynomial both vanish; solve the resulting linear + quadratic conditions for the two direction ratios.

Solution

Step 1 — The point lies on the surface; the surface is a ruled quadric

At $(1,1,1)$: $(1+1+1)(2+1-1)=3\cdot2=6=6(1)$. ✓ The equation

expands to a quadratic in $x,y,z$ whose quadratic part is the product of two linear forms — this is a non-central (ruled) quadric (a hyperbolic paraboloid), so through each of its points pass two straight-line generators.

Step 2 — Impose that a line through $(1,1,1)$ lies on $S$

Take a line $(x,y,z)=(1+a t,\,1+b t,\,1+c t)$ with direction $(a,b,c)$. Substitute and expand $S$ as a polynomial in $t$:

For the entire line to lie on $S$, both the $t^1$ and $t^2$ coefficients must vanish:

Step 3 — Solve for the two directions

From (L), $c=\dfrac{8a+5b}{7}$. Substitute into (Q) and clear $49$:

So either $b=-3a$ or $b=-\tfrac54 a$:

• $b=-3a$: take $a=1\Rightarrow b=-3,\ c=\dfrac{8-15}{7}=-1$. Direction $(1,-3,-1)$.
• $b=-\tfrac54a$: take $a=4\Rightarrow b=-5,\ c=\dfrac{32-25}{7}=1$. Direction $(4,-5,1)$.

Step 4 — The two generators

Answer