Question

If the line $lx+my+n=0$ meets the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the extermities of a pair of conjugate diameters, then

• A. $a^2{l}^2-b^2{m}^2=0$
• B. $a^2{l}^2-b^2{m}^2=1$
• C. $a^2{l}^2-b^2{m}^2=2$
• D. $a^2{l}^2-b^2{m}^2=3$

Answer 1

The correct option is A $a^2{l}^2-b^2{m}^2=0$

Let $\theta$ and $\varphi$ be the essentric angles of the conjugate diameters,
Then,
$\theta +\varphi =\frac{\pi }{2}$
$(a\sec \theta ,b\tan \theta ),(a\sec \varphi ,b\tan \varphi )$
$\Rightarrow (a\sec \theta ,b\tan \theta ),(a\text{cosec}\theta ,b\cot \theta )$

$(y-b\tan \theta )=\frac{b\cot \theta -b\tan \theta }{a\text{cosec}\theta -a\sec \theta }(x-a\sec \theta )$

$\Rightarrow ya-ab\tan \theta =xb(\sin \theta +\cos \theta )-ab(\tan \theta +1)$
$ya=xb(\sin \theta +\cos \theta )-ab$
$ya-xb(\sin \theta +\cot \theta )+ab=0$
$lx+my+n=0$

Comparing both, we get

$\frac{a}{m}=\frac{-b(\sin \theta +\cos \theta )}{l}=\frac{ab}{n}$

On elliminating $\theta$, we can get
$a^2{l}^2-b^2{m}^2=0$