Question

If the normal at a point P to the hyperbola meets the transverse axis at G, and the value of $\frac{SG}{SP}$ is 2 then the eccentricity of the hyperbola is (where S is the focus of the hyperbola)

Answer 1

Given hyperbola is,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Let

$P\left(\theta \right)\equiv (asec\theta ,btan\theta )$

Equation of normal at $P\theta$ is given by,

$acos\theta .x+b.cot\theta .y=a^2+b^2$

This normal meets transverse axis at $G(x_1,0)$

$\therefore acos\theta x_1+0=a^2+b^2$

$\therefore G\equiv [\frac{a^2+b^2}{acos\theta },0]$

Focus $S\equiv (ae,0)$

$SP=\sqrt{a^2(e-sec\theta {)}^2+(btan\theta {)}^2}$

$=\sqrt{a^2{e}^2+a^{2}se{c}^2\theta -2a^2.esec\theta +b^{2}ta{n}^2\theta }$

$=\sqrt{a^2+b^2+a^{2}se{c}^2\theta -2a^2.esec\theta +b^{2}ta{n}^2\theta }$

$=\sqrt{a^2+a^{2}se{c}^2\theta -2a^2.esec\theta +se{c}^2\theta .a^2(e^2-1)}$

$=\sqrt{a^2-2a^{2}esec\theta +se{c}^2\theta a^2.e^2}$

$\sqrt{(e.a.sec\theta -a{)}^2}$

$=easec\theta -a$

$\therefore SP=e.asec\theta -a$

$SG=\frac{a^2+b^2}{acos\theta }-ae=\frac{a^2{e}^2}{acos\theta }-ae(Since{a}^2{e}^2=a^2+b^2)$

$=a{e}^2.sec\theta -ae$

$=e[aesec\theta -a]$

$\therefore \frac{SG}{SP}=\frac{e.(aesec\theta -a)}{(aesec\theta -a)}=e=2$