Question

If latus rectum of a hyperbola subtends a right at other focus of hyperbola, then is eccentricity is equal to-

• A. $2$
• B. $\tan \frac{\pi }{8}$
• C. $\cot \frac{\pi }{8}$
• D. $\left(\frac{1}{\sqrt{2}-1}\right)$

Answer 1

The correct option is A $2$

Equation of parabola is-

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

At $(ae,y_1)$,

$\frac{{\left(ae\right)}^2}{a^2}-\frac{{y_1}^2}{b^2}=1$

$\Rightarrow \frac{{y_1}^2}{b^2}=e^2-1$

$\Rightarrow y_1=\pm b\sqrt{e^2-1}$

As $AB\perp BC$,

$\text{slope of AB}\times \text{slope of BC}=-1⟶\left(1\right)$

Now,

$\text{slope of AB}=\frac{b\sqrt{e^2-1}}{ae+a}$

$\text{slope of BC}=\frac{-b\sqrt{e^2-1}}{ae+a}$

$\therefore$ Substituting these values in ${eq}^n\left(1\right)$, we have

$\frac{b\sqrt{e^2-1}}{ae+a}\times \frac{-b\sqrt{e^2-1}}{ae+a}=-1$

$\Rightarrow b^2(e^2-1)={(ae+a)}^2$

$\Rightarrow a^2(e^2-1)(e^2-1)=a^2{(e+1)}^2\{\because b^2=a^2(e^2-1)\}$

$\Rightarrow e^4+1-2e^2=e^2+1+2e$

$\Rightarrow e^4-3e^2-2e=0$

$\Rightarrow e(e^3-4e-2)=0$

$\Rightarrow e(e-2){(e+1)}^2=0$

$\therefore e=0\text{or}e-2=0\text{or}e+1=0$

$\Rightarrow e=0\text{or}e=2\text{or}e=-1$

Since $e>1$ for hyperbola,

$\therefore e=2$