Question

Tangent to the ellipse \(\frac{x^2}{32} + \frac{y^2}{18} = 1\) having slope \(-\frac{3}{4}\) meet the coordinate axis at A and B. Then, the area of \(\triangle AOB\), where O is the origin, is

• A. 12 sq units
• B. 8 sq units
• C. 24 sq units
• D. 32 sq units

Hint:

For ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), tangent with slope \(m\) is \(y = mx \pm \sqrt{a^2 m^2 + b^2}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Equation of tangent to ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with slope \(m\) is \(y = mx \pm \sqrt{a^2 m^2 + b^2}\).

Step 2: Detailed Explanation:
Here \(a^2 = 32\), \(b^2 = 18\), \(m = -\frac{3}{4}\).
\(\sqrt{a^2 m^2 + b^2} = \sqrt{32 \times \frac{9}{16} + 18} = \sqrt{18 + 18} = \sqrt{36} = 6\).
Equation: \(y = -\frac{3}{4}x \pm 6\).
Taking positive: \(y = -\frac{3}{4}x + 6\).
Intercepts: \(x\)-intercept: \(0 = -\frac{3}{4}x + 6 \implies x = 8\).
\(y\)-intercept: \(y = 6\).
Area = \(\frac{1}{2} \times 8 \times 6 = 24\).

Step 3: Final Answer:
Option (C) 24 sq units.