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 \[\left( ae, \pm y \right)\]. 

\[\therefore \frac{\left( ae \right)^2}{a^2} - \frac{y^2}{b^2} = 1\]

\[ \Rightarrow \frac{y^2}{b^2} = \frac{\left( ae \right)^2}{a^2} - 1\]

\[ \Rightarrow \frac{y^2}{b^2} = e^2 - 1\]

\[ \Rightarrow y = \pm b\sqrt{e^2 - 1}\]

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  \[- 1\].

\[\Rightarrow \frac{b\sqrt{e^2 - 1}}{ae + a} \times \frac{- b\sqrt{e^2 - 1}}{ae + a} = - 1\]

\[ \Rightarrow b^2 \left( e^2 - 1 \right) = \left( ae + a \right)^2 \]

\[ \Rightarrow b^2 \left( e^2 - 1 \right) = a^2 \left( e + 1 \right)^2 \]

\[ \Rightarrow \left( e^2 - 1 \right)^2 = \left( e + 1 \right)^2 \]

\[ \Rightarrow e^4 - 2 e^2 + 1 = e^2 + 2e + 1\]

\[ \Rightarrow e^4 - 3 e^2 - 2e = 0\]

\[ \Rightarrow e\left( e^3 - 3e - 2 \right) = 0\]

\[ \Rightarrow e\left( e - 2 \right)\left( e + 1 \right)\left( e + 1 \right) = 0\]

∴ \[e = 0, - 1, 2\]

The value of e can neither be negative nor zero.
Therefore, the value of eccentricity is 2.