Question

If the latus rectum through one focus subtends a right angle at the farther vertex of the hyperbola then its eccentricity is:

• A. $\sqrt{2}$
• B. $2$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $2$

Let the equation of Hyperbola : $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The Latus Rectum $L{L}^{{}^{\prime }}$ is substending ${90}^{\circ }$at $A^{{}^{\prime }}$. So, the lines $A^{{}^{\prime }}L$, $A^{{}^{\prime }}{L}^{{}^{\prime }}$ and $L{L}^{{}^{\prime }}$ are sides of right angled triangle.

So, we can apply Pythagoras Theorem to it.

where $L{L}^{{}^{\prime }}=\frac{2b^2}{a}=$Length of LR

$A^{{}^{\prime }}L=\sqrt{(ae+a{)}^2+(\frac{b^2}{a}-0{)}^2}$

$A^{{}^{\prime }}{L}^{{}^{\prime }}=A^{{}^{\prime }}L$(due to symmetry about X-axis)

So, by Pythagoras Theorem,

$(A^{{}^{\prime }}{L}^{{}^{\prime }}{)}^2+(A^{{}^{\prime }}L{)}^2=(L{L}^{{}^{\prime }}{)}^2$

$\Rightarrow 2\left(\right(ae+a{)}^2+(\frac{b^2}{a}-0{)}^2)=\frac{4b^4}{a^2}$

$\Rightarrow 2\left(a^2\right(e+1{)}^2+\frac{b^4}{a^2})=\frac{4b^2}{a^2}$

$\Rightarrow 2a^2(e+1{)}^2=\frac{b^4}{a^4}$

$\Rightarrow (e+1{)}^2=\frac{b^4}{a^4}$ (as $b^2=a^2(e^2-1))$

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

$\Rightarrow e+1=\pm (e^2-1)$

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

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

$\Rightarrow e^2-2e+1e-2=0$

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

$\Rightarrow e\ne -1,\Rightarrow e=2$