Question

Find the area of the triangle formed by the tangent to the curve \( xy = a^2 \) at \( (x_1, y_1) \) and the coordinate axes.

• A. \( a^2 \)
• B. \( \frac{3a^2}{2} \)
• C. \( 2a^2 \)
• D. \( 4a^2 \)

Hint:

To find triangle area formed by a tangent and coordinate axes, use intercept form after finding the tangent.

Answer 1

The Correct Option is C

Given: \( xy = a^2 \Rightarrow y = \frac{a^2}{x} \) Differentiate: \[ \begin{align} \frac{dy}{dx} = -\frac{a^2}{x^2} \Rightarrow \text{slope at } (x_1, y_1) = -\frac{y_1}{x_1} \] Equation of tangent: \[ \begin{align} y - y_1 = -\frac{y_1}{x_1}(x - x_1) \Rightarrow xy_1 + x_1 y = 2a^2 \] To find intercepts:
- Put \( y = 0 \Rightarrow x = \frac{2a^2}{y_1} \)
- Put \( x = 0 \Rightarrow y = \frac{2a^2}{x_1} \)
So intercepts on axes are:
- x-intercept = \( \frac{2a^2}{y_1} \)
- y-intercept = \( \frac{2a^2}{x_1} \)
Area of triangle: \[ \begin{align} \text{Area} = \frac{1}{2} \cdot \frac{2a^2}{x_1} \cdot \frac{2a^2}{y_1} = \frac{2a^4}{x_1 y_1} \text{ and since } x_1 y_1 = a^2 \Rightarrow \text{Area} = 2a^2 \]