If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\sec \alpha$, then area of the…

If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\sec \alpha$, then area of the triangle formed by the asymptotes of the hyperbola with any of its tangent is
  1. $a^2 b^2 \sec ^2 \alpha$
  2. $\frac{b^2}{|\tan \alpha|}$
  3. $a^2 \tan ^2 \alpha$
  4. $\left(a^2+b^2\right) \tan ^2 \alpha$

Solution

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
We know, area of triangle formed by the asymptotes of the hyperbola with any of its tangent is $a b$. Now, $e^2=\sec ^2 \alpha$ $\begin{aligned} & \Rightarrow 1+\frac{b^2}{a^2}=1+\tan ^2 \alpha \Rightarrow \frac{b^2}{a^2}=\tan ^2 \alpha \\ & \Rightarrow \frac{b^2}{\tan ^2 \alpha}=a^2 \Rightarrow \frac{b^4}{\tan ^2 \alpha}=a^2 b^2 \\ & \Rightarrow \frac{b^2}{|\tan \alpha|}=a b \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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