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

The equation of a hyperbola is given by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). The eccentricity \( e \) of a hyperbola is defined as \( e = \sqrt{1+\frac{b^2}{a^2}} \). Given that \( e = \sec \alpha \), we equate the two expressions for eccentricity:
\[ \sec \alpha = \sqrt{1+\frac{b^2}{a^2}} \] Squaring both sides, we have:
\[ \sec^2 \alpha = 1+\frac{b^2}{a^2} \] Therefore,
\[ \frac{b^2}{a^2} = \sec^2 \alpha - 1 \]
Using the identity \(\sec^2 \alpha - 1 = \tan^2 \alpha\), it follows that:
\[ \frac{b^2}{a^2} = \tan^2 \alpha \] Now, consider the equations of the asymptotes for the given hyperbola, which are:
\[ y = \pm \frac{b}{a}x \] The tangent equation \(\frac{x_1x}{a^2} - \frac{y_1y}{b^2} = 1\) intersects these asymptotes at a triangle whose area we need to find. The area \( A \) of this triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x_1 \times y_1 \] where \( x_1 = \frac{a^2}{\sqrt{a^2+b^2\tan^2\alpha}} \) and \( y_1 = \frac{b^2}{\sqrt{a^2+b^2\tan^2\alpha}} \). Substituting \( \frac{b^2}{a^2} = \tan^2 \alpha\) into:
\(\sqrt{a^2+b^2\tan^2\alpha} = \sqrt{a^2 + b^2 \frac{b^2}{a^2}}\), simplifies to \( a\sec\alpha\). Hence:
\[ A = \frac{1}{2} \times \frac{a^2}{a\sec\alpha} \times \frac{b^2}{b\sec\alpha} = \frac{1}{2} \times \frac{a^2}{a} \times \frac{b^2}{b} \times \sin^2\alpha \] Considering that \(\sin\alpha = \frac{1}{\sec\alpha}\) or \( \sin\alpha = \frac{1}{\sqrt{1+\tan^2\alpha}}\), the area simplifies to
\[ A = \frac{b^2}{2|\tan\alpha|}\] Therefore, the area of the triangle formed is:
\(\frac{b^2}{|\tan \alpha|}\)

Answer 2

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 

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