Question

If the normal at a point $P$ to the hyperbola meets the transverse axis at $G$ and the value of $\frac{S G}{S P}$ is $6$ , then the eccentricity of the hyperbola is (where $S$ is the focus of hyperbola)

Answer 1

Let the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ .
Focus $S\left(a e , \, 0\right)$
Now, equation of normal at the parametric point $P\left(a \sec \theta , b \tan \theta \right)$ is $a\cos\theta x+b\cot\theta y=a^{2}+b^{2}$

Now it intersects the transverse axis i.e., $x$ -axis at $G$ . So put $y=0$ in the equation of the normal, $$
$\therefore G\left(\frac{a^{2} + b^{2}}{a \cos \theta } , 0\right)$
Focal distance, $SP=e\left(a \sec \theta \right)-a...\left(1\right)$
Since $S$ and $G$ lie on $x$ -axis, we can directly write,
$SG=\frac{a^{2} + b^{2}}{a \cos \theta }-ae=\frac{a^{2} e^{2}}{a \cos \theta }-ae=e\left[e \left(a \sec \theta \right) - a\right]...\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ ,
$\Rightarrow \frac{S G}{S P}=e=6$