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 $6$ , then the eccentricity of the hyperbola is (where $S$ is the focus of hyperbola)

Answer 1

Answer: 6

Let the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .
Focus $S\left(ae,0\right)$
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 $\left(2\right)$ ,
$\Rightarrow \frac{SG}{SP}=e=6$