Question

If the line $lx+my+n=0$ is a norma to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then show that $\frac{a^2}{l^2}-\frac{b^2}{m^2}={\left(\frac{a^2+b^2}{n}\right)}^2$

Answer 1

equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

equation of normal $\frac{ax}{sec\theta }+\frac{by}{tan\theta }=a^2+b^2$ ______ (1)

lx+my+b=0 ______ (2) is normal to the parabola

$\Rightarrow \frac{l}{a/sec\theta }=\frac{m}{b/tan\theta }=\frac{n}{-(a^2+b^2)}$

$\Rightarrow \frac{lsec\theta }{a}=\frac{mtan\theta }{b}=\frac{-n}{a^2+b^2}$

$\Rightarrow sec\theta =\frac{-an}{l(a^2+b^2)}$ $tan\theta \theta =\frac{-bn}{m(a^2+b^2)}$

We ahve $se{c}^2\theta -ta{n}^2\theta =1$

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

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

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