Q: The line lx + my + n = 0 will be a normal to the hyperbola b^{2} x^{2} – a^{2} y^{2} = a^{2} b^{2} if

(A) $\large \frac{a^2}{l^2} + \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2} $

(B) $\large \frac{a^2}{l^2} – \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2} $

(C) $\large \frac{a^2}{l^2} – \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n} $

(D) none of these

Sol. Equation of normal at (asecq, btanq) is;

ax cosq + by cotq = a2 +b2

Comparing it with lx+my + n = 0 we get

$\large \frac{a cos\theta}{l} = \frac{b cot\theta}{m} = \frac{a^2 + b^2 }{-n}$

$\large cos\theta = \frac{l(a^2 + b^2)}{-n a}$ and $\large cot\theta = \frac{m(a^2 + b^2)}{-n b}$

$\large sin\theta = \frac{b l}{a m}$

Thus , $\large \frac{b^2 l^2}{a^2 m^2} + \frac{l^2(a^2 + b^2)^2}{a^2 n^2} = 1$

or , $\large \frac{a^2}{l^2} – \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2} $

Hence (B) is the correct answer.