Question

If the eccentricity of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is \( \sec \alpha \), then the area of the triangle formed by the asymptotes of the hyperbola with any of its tangent is:

• A. \( a^2b^2 \sec^2\alpha \)
• B. \( \frac{b^2}{|\tan \alpha|} \)
• C. \( a^2\tan^2\alpha \)
• D. \( (a^2+b^2)\tan^2\alpha \)

Hint:

For hyperbolas, the asymptotes play an important role in defining the region where a tangent interacts. The area of the triangle formed by asymptotes and a tangent is calculated using the semi-axis values.

Answer 1

The Correct Option is B

We are given the hyperbola: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] The eccentricity of the hyperbola is \( e = \sec \alpha \). Step 1: Relating Eccentricity and Hyperbola Parameters For a hyperbola, the eccentricity is given by: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] Since \( e = \sec \alpha \), we have: \[ \sqrt{1 + \frac{b^2}{a^2}} = \sec \alpha \] Squaring both sides, \[ 1 + \frac{b^2}{a^2} = \sec^2 \alpha \] Since \( \sec^2 \alpha = 1 + \tan^2 \alpha \), we can substitute this identity: \[ 1 + \frac{b^2}{a^2} = 1 + \tan^2 \alpha \] \[ \frac{b^2}{a^2} = \tan^2 \alpha \] From this, \[ b^2 = a^2 \tan^2 \alpha \] Step 2: Area of Triangle Formed by Asymptotes and a Tangent The area of the triangle formed by the asymptotes with any tangent is given by: \[ \text{Area} = \frac{b^2}{| \text{Slope of the Asymptote} |} \] The slope of the asymptote is \( \frac{b}{a} = \tan \alpha \). Thus, \[ \text{Area} = \frac{b^2}{|\tan \alpha|} \] Step 3: Final Answer 

\[Correct Answer: (2) \ \frac{b^2}{|\tan \alpha|}\]