Question

If the latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex, then write the eccentricity of the hyperbola.

Answer 1

The standard equation of hyperbola is given below:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Let the latus rectum be QR, which passes through one of the focus. QR subtends a right angle at point P.

It passes through the focus $\left(ae,\pm y\right)$.
$$\begin{gathered} \therefore \frac{{\left(ae\right)}^2}{a^2}-\frac{y^2}{b^2}=1 \\ \Rightarrow \frac{y^2}{b^2}=\frac{{\left(ae\right)}^2}{a^2}-1 \\ \Rightarrow \frac{y^2}{b^2}=e^2-1 \\ \Rightarrow y=\pm b\sqrt{e^2-1} \end{gathered}$$
Now, the slope of PQ is $\frac{b\sqrt{e^2-1}}{ae+a}$ and the slope of PR is$-\frac{b\sqrt{e^2-1}}{ae+a}$.
The lines PQ and PR are perpendicular, so the product of the slope is $-1$.

$$\begin{gathered} \Rightarrow \frac{b\sqrt{e^2-1}}{ae+a}\times \frac{-b\sqrt{e^2-1}}{ae+a}=-1 \\ \Rightarrow b^2\left(e^2-1\right)={\left(ae+a\right)}^2 \\ \Rightarrow b^2\left(e^2-1\right)=a^2{\left(e+1\right)}^2 \\ \Rightarrow {\left(e^2-1\right)}^2={\left(e+1\right)}^2 \\ \Rightarrow e^4-2e^2+1=e^2+2e+1 \\ \Rightarrow e^4-3e^2-2e=0 \\ \Rightarrow e\left(e^3-3e-2\right)=0 \\ \Rightarrow e\left(e-2\right)\left(e+1\right)\left(e+1\right)=0 \end{gathered}$$

∴ $e=0,-1,2$

The value of e can neither be negative nor zero.

Therefore, the value of eccentricity is 2.