Question

A hyperbola passes through $(1,2)$ and has asymptotes as $x+y-2=0$ and $7x+y-8=0$. Eccentricity of hyperbola is equal to

• A. $\frac{\sqrt{10}}{3}$
• B. $\sqrt{1}0$
• C. $\frac{\sqrt{7}}{2}$
• D. $\frac{\sqrt{7}}{3}$

Answer 1

The correct option is A $\frac{\sqrt{10}}{3}$

We know that 'a' length of transverse axis, 'b' length of conjugate axis.

Acute angle between asymptotes is given by $2{\tan }^{-1}\left(b/a\right)$

Let $\varphi$ be angle between asymptotes

Given $x+y=2$ & $7x+y=8$

$\tan \varphi =|-1+7|/1+7=6/8=3/4$

As $\varphi =2{\tan }^{-1}\left(b/a\right)$

$\tan \varphi /2=b/a=x$.

$\tan \varphi =2\tan \left(\varphi /2\right)/1-{\tan }^2\left(\varphi /2\right)$

$\frac{3}{4}=2x/1-x^2$

$3-3x^2=8x$

$3x^2+8x-3=0$

$(3x-1)(x+3)=0$

$x=1/3$ or $-3$.

Hence $b/a=1/3$

Eccentricity of hyperbola:-

$e=\sqrt{1+(b/a{)}^2}=\sqrt{1+1/9}=\frac{\sqrt{10}}{3}$.