Question

If \(P(\alpha, \beta)\) is a point on the curve \(9x^2 + 4 y^2 = 144\) in the first quadrant and the minimum area of the triangle formed by the tangent of the curve at \(P\) with the coordinate axes is \(S\), then find \(S\).

• A. \(S = \sqrt{\alpha \beta}\)
• B. \(S = \alpha \beta\)
• C. \(S = 2 \sqrt{\alpha \beta}\)
• D. \(S = 2 \alpha \beta\)

Hint:

Use parametric form of ellipse and area formula for tangent intercept triangle.

Answer 1

The Correct Option is D

Equation: \(\frac{x^2}{16} + \frac{y^2}{36} = 1\) (dividing both sides by 144). Tangent line intercepts with axes form triangle of area \[ S = \frac{1}{2} (x\text{-intercept}) (y\text{-intercept}). \] Using slope form and minimizing \(S\), minimum area is \[ S = 2 \alpha \beta. \]