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UPSC 2026 Maths Optional Paper 1 Q4c-i — Step-by-Step Solution

10 marks · Section A

Hyperboloid of one sheet · Analytic Geometry · asked 3× in 14 yrs · Read the full method →

Question

Find the equations of the generating lines of the hyperboloid

which pass through the points $(2, 3, -4)$ and $\left(2, -1, \dfrac{4}{3}\right)$.

Technique

The surface is a hyperboloid of one sheet, a doubly ruled surface. Factorise it as $\left(\frac x2-\frac z4\right)\left(\frac x2+\frac z4\right)=\left(1-\frac y3\right)\left(1+\frac y3\right)$ and split it into the two standard one-parameter systems of generators ($\lambda$-system and $\mu$-system). Through each point of the surface passes exactly one generator of each system; substituting the point fixes the parameter. Handle the degenerate member $\lambda=\infty$ by writing the system with a homogeneous parameter $(l:m)$.

Solution

Step 1 — Verify that both points lie on the surface.

This must be done first: a point off the surface has no generator through it.

For $P_1(2,3,-4)$: $\dfrac{4}{4}+\dfrac{9}{9}-\dfrac{16}{16}=1+1-1=1.\ \checkmark$

For $P_2\!\left(2,-1,\dfrac43\right)$: $\dfrac{4}{4}+\dfrac{1}{9}-\dfrac{(4/3)^2}{16}=1+\dfrac19-\dfrac{16/9}{16}=1+\dfrac19-\dfrac19=1.\ \checkmark$

Both lie on the hyperboloid.

Step 2 — The two systems of generators.

Rearrange:

Since $\dfrac{x^2}{4}+\dfrac{y^2}{9}-\dfrac{z^2}{16}=1$ is a hyperboloid of one sheet, it is doubly ruled. The two systems are:

Multiplying the two equations of either system reproduces $(1)$ identically (the parameter cancels), so every member of either system lies wholly on the surface. Each is a one-parameter family of lines (intersection of two planes), and through any point of the surface passes exactly one line of each family.

To include the limiting member $\lambda=\infty$, write System I with a homogeneous parameter $(l:m)\ne(0:0)$:

which reduces to (I) with $\lambda=m/l$ when $l\ne 0$, and gives the degenerate generator $\left\{1-\frac y3=0,\ \frac x2+\frac z4=0\right\}$ when $l=0$.

Step 3 — Generators through $P_1(2,3,-4)$.

At $P_1$:

System I. From (I$'$): $l\cdot 2=m\cdot 0\Rightarrow l=0$, hence $m\ne0$ and the generator is

i.e.

(This is the $\lambda\to\infty$ member; note $y=3$ makes the right side of (1) vanish, which is exactly why no finite $\lambda$ works.) Its direction ratios are $(0,1,0)\times(2,0,1)=(1,0,-2)$, so in symmetric form

System II. From (II): $2=\mu\cdot 2\Rightarrow \mu=1$, and the second equation checks: $1\cdot 0=0=1-\frac y3.\ \checkmark$ The generator is

Multiplying each by $12$: $6x-4y-3z=12$ and $6x+4y+3z=12$. Adding gives $12x=24$, subtracting gives $8y+6z=0$; hence

Direction ratios $(1,0,0)\times(0,4,3)=(0,-3,4)$, i.e. $(0,3,-4)$, so

Step 4 — Generators through $P_2\!\left(2,-1,\tfrac43\right)$.

At $P_2$:

System I. $\dfrac23=\lambda\cdot\dfrac43\Rightarrow \lambda=\dfrac12$; check the second equation: $\dfrac12\cdot\dfrac43=\dfrac23=1+\dfrac y3.\ \checkmark$ The generator is

Clearing fractions ($\times12$ and $\times24$ respectively):

(Check at $P_2$: $12-2-4=6\ \checkmark$; $12+8+4=24\ \checkmark$.) Direction ratios

so

System II. $\dfrac23=\mu\cdot\dfrac23\Rightarrow \mu=1$; check: $1\cdot\dfrac43=\dfrac43=1-\dfrac y3.\ \checkmark$ But $\mu=1$ is the same value found at $P_1$, so this generator is again

which indeed passes through $P_2$: $x=2\ \checkmark$ and $4(-1)+3\left(\tfrac43\right)=-4+4=0\ \checkmark$.

Step 5 — Collect the answer, and note the coincidence.

Through $P_1$: $\ell_1$ (System I) and $\ell_2$ (System II). Through $P_2$: $\ell_3$ (System I) and $\ell_2$ (System II).

So the four generators asked for are only three distinct lines: $P_1$ and $P_2$ lie on a common generator $\ell_2$. This is visible at once from the data — both points have $x=2$, and $4(3)+3(-4)=0$, $4(-1)+3\left(\tfrac43\right)=0$ — so the whole line joining them, $x=2,\ 4y+3z=0$, is a ruling. (Consistently, the $\mu$-value came out as $1$ at both points.)

Step 6 — Confirm each line lies wholly on the surface.

Substitute a parametrisation into $S(x,y,z)=\dfrac{x^2}{4}+\dfrac{y^2}{9}-\dfrac{z^2}{16}$ and check it is identically $1$ in $t$ (not merely at one point).

$\ell_1$: $(x,y,z)=(2+t,\,3,\,-4-2t)$.

$\ell_2$: $(x,y,z)=(2,\,3+3t,\,-4-4t)$.

(Setting $t=-\tfrac43$ gives $(2,-1,\tfrac43)=P_2$, confirming $\ell_2$ meets $P_2$.)

$\ell_3$: $(x,y,z)=\left(2+3t,\,-1+6t,\,\tfrac43+10t\right)$.

Therefore

All three lines lie wholly on the hyperboloid.

Answer