Question

If the latus rectum subtends a right angle at the center of the hyperbola then its eccentricity.

• A. $e=\left(\sqrt{13}\right)/2$
• B. $e=\left(\sqrt{5}-1\right)/2$
• C. $e=\left(\sqrt{5}+1\right)/2$
• D. $e=\left(\sqrt{3}+1\right)/2$

Answer 1

Answer: (B), (C)

The correct options are
B $e=\left(\sqrt{5}-1\right)/2$
C $e=\left(\sqrt{5}+1\right)/2$

Given $LR$ of hyperbola subtends ${90}^0$ at its centre

Let there be a hyperbola : $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$----------------(1)

So, eccentricity , e = $\sqrt{1+\frac{b^2}{a^2}}$

Ends of $LR$, $L$: $\left(ae,\frac{b^2}{a}\right)$ &L'$\left(ae,\frac{-b^2}{a}\right)$

$C$ : centre$(0,0)$

$\angle LC{L}^{\prime }={90}^0$

$\Rightarrow △LC{L}^{\prime }=$right angled triangle

$\Rightarrow$ It must follow pythogores theorem ($H^2=P^2{+B}^2$)

For$LC$ and $L^{\prime }C$and $L{L}^{\prime }$

$LL$'=length of $\frac{2b^2}{a}$

$\overline{LC}=\sqrt{(ae-0{)}^2+(\frac{b^2}{a}-0)}=\sqrt{a^2{e}^2+\frac{b^4}{a^2}}=\sqrt{\frac{a^4{e}^2+b^4}{a^2}}$--------------------(2)

$\overline{LC}=\overline{L^{\prime }C}$ due to symmetry along x axis--------------(3)

$\Rightarrow$ Applying Pythagoras theorem , ${\left(\overline{L{L}^{\prime }}\right)}^2={\left(\overline{LC}\right)}^2{\left(\overline{L^{\prime }C}\right)}^2$

$\Rightarrow \frac{4b^4}{a^2}=2(LC{)}^2$ from equation (3)

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

$\Rightarrow \frac{4b^4}{a^2}-\frac{2b^4}{a^2}=2a^2{e}^2$

$\Rightarrow \frac{2b^4}{a^2}=2a^2{e}^2$

$\Rightarrow b^4=a^4{e}^2$

$\Rightarrow {(a^2{(e}^2-1))}^2=a^4{e}^2$

$(a^2{(e}^2-1))$

$\Rightarrow a^4{(e}^2-1{)}^2=a^4{e}^2$

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

$e=\frac{\pm 1\pm \sqrt{1+4}}{2}=\frac{\pm 1\pm \sqrt{5}}{2}$

$e\ne \frac{-1-\sqrt{5}}{2}$ (negative)$,\frac{1-\sqrt{5}}{2}$

$e=\frac{\sqrt{5}-1}{2},\frac{1+\sqrt{5}}{2}$.