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UPSC Maths 2013 Paper 1 Q4c — Solution

20 marks · Section A

Question

A variable generator meets two generators of the system through the extremities $B$ and $B'$ of the minor axis of the principal elliptic section of the hyperboloid $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}-z^{2}c^{2}=1$ in $P$ and $P'$. Prove that $BP\cdot B'P'=a^{2}+c^{2}$.

(Note: the PDF writes "$z^{2}c^{2}$"; the standard reading is $z^{2}/c^{2}$ — i.e. the standard hyperboloid of one sheet. We adopt the standard form.)

Technique

Standard ruled-surface parametrisation; the two families of generators of a hyperboloid of one sheet; explicit parametric intersection calculations.

Solution

Strategy. Identify the two ruled-line families of the hyperboloid; pick the two “fixed” generators through $B$ and $B'$ in one family; the variable generator from the other family meets both, with intersection points $P,P'$. Compute distances.

Step 1 — Two ruled families

Rewrite the hyperboloid as

This yields two families of lines (rulings):

Family $\Lambda$ (parameter $\lambda$):

Family $M$ (parameter $\mu$):

Step 2 — Generators through $B$ and $B'$

$B=(0,b,0),\;B'=(0,-b,0)$ are endpoints of the minor axis of the principal ellipse (which is the $z=0$ section, the ellipse $x^{2}/a^{2}+y^{2}/b^{2}=1$; minor axis is along $y$ when $a>b$, with endpoints $(0,\pm b,0)$).

At $B$ (substitute $y=b$, $1+y/b=2$, $1-y/b=0$): family $M$ with $\mu\to 0$ gives the line $\{y=b,\;x/a=z/c\}$. Direction $(a,0,c)$. Call this $L_B$.

At $B'$ (substitute $y=-b$, $1+y/b=0$, $1-y/b=2$): family $M$ with $\mu\to\infty$ gives the line $\{y=-b,\;x/a=-z/c\}$. Direction $(a,0,-c)$. Call this $L_{B'}$.

Both $L_B$ and $L_{B'}$ are in family $M$ — “two generators of the same system through $B$ and $B'$.” They are skew (different $y$-values, not parallel — direction $(a,0,c)$ vs $(a,0,-c)$).

Step 3 — Variable generator from family $\Lambda$ meets both

Take $\ell_\lambda$, the family-$\Lambda$ generator at parameter $\lambda$. We compute its intersection with $L_B$ and $L_{B'}$.

Intersection $P=\ell_\lambda\cap L_B$: at $y=b$, so $1-y/b=0,\;1+y/b=2$. Plug into the family-$\Lambda$ equations:

So $x/a=z/c=1/\lambda$, giving

Intersection $P'=\ell_\lambda\cap L_{B'}$: at $y=-b$, so $1-y/b=2,\;1+y/b=0$. Plug into family-$\Lambda$:

So $x/a=\lambda,\;z/c=-\lambda$, giving

Step 4 — Compute $BP$ and $B'P'$

$\overrightarrow{BP}=P-B=\bigl(a/\lambda,\;0,\;c/\lambda\bigr)$.

$\overrightarrow{B'P'}=P'-B'=(a\lambda,\;0,\;-c\lambda)$.

Step 5 — Product

Answer