Question

The normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ drawn at an extremity of its latus rectum is parallel to an asymptote. Show that eccentricity is equal to the square root of $\frac{(1+\sqrt{5})}{2}$.

Answer 1

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

To prove Eccentricity,

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

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

$\frac{x}{a}\pm \frac{y}{b}=0$ - Equation 1

Considering normal is drawn at L, it must be parallel to asymptote $\frac{x}{a}+\frac{y}{b}=0$

and Equation of normal at $L(ae,\frac{b^2}{a})$ is

$N:\frac{a^{2}x}{ae}+\frac{b^{2}y}{\frac{b^2}{a}}=a^2+b^2$

$\Rightarrow \frac{ax}{e}+ay=a^2+b^2$ - Equation 2

As Equation 2 is parrallel to $\frac{x}{a}+\frac{y}{b}=0$, their slopes must be equal.

Slope of $\frac{x}{a}+\frac{y}{b}=0,m_1=\frac{b}{a}$

Slope of $\frac{ax}{e}+ay=a^2+b^2\Rightarrow y=-\frac{1}{e}x+\frac{a^2+b^2}{a}$

Slope $m_2=-\frac{1}{e}$

As $m_1=m_2\Rightarrow \frac{b}{a}=-\frac{1}{e}$

$\frac{b^2}{a^2}=\frac{1}{e^2}$ - Equation 3

We know $b^2=a^2(e^2-1)$

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

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

$\Rightarrow e^4-e^2-1=0$

Let's take $e^2=t$

$\Rightarrow t^2-t-1=0$

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

It cannot be $\frac{1-\sqrt{5}}{2}$ as for Hyperbola $e>1,e^2>1$

$\Rightarrow t=e^2=\frac{1+\sqrt{5}}{2}$

$\Rightarrow e=\pm \sqrt{\frac{1+\sqrt{5}}{2}}$ again $e\ne negative\Rightarrow e=\sqrt{\frac{1+\sqrt{5}}{2}}$