Question

The area of the triangle formed by the co-ordinate axes and a tangent to the curve $xy = a^2$ at the point $(x_1, y_1)$ is ______ sq. units (where $a, x_1$ and $y_1$ are non-zero)

• A. $\frac{a^2 x_1}{y_1}$
• B. $\frac{a^2 y_1}{x_1}$
• C. $2a^2$
• D. $4a^2$

Hint:

For any rectangular hyperbola $xy = c$, the area of the triangle formed by the tangent and axes is always $2c$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the area of the triangle formed by the axes and the tangent line to a hyperbola at a specific point.
Step 2: Detailed Explanation:
1. Curve: \( xy = a^2 \). Differentiating: \( x \frac{dy}{dx} + y = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \).
2. At point \( (x_1, y_1) \), slope \( m = -\frac{y_1}{x_1} \).
3. Equation of tangent: \( y - y_1 = -\frac{y_1}{x_1} (x - x_1) \).
\[ x_1 y - x_1 y_1 = -y_1 x + x_1 y_1 \]
\[ y_1 x + x_1 y = 2 x_1 y_1 \]
4. Intercepts of the tangent:
- X-intercept (put \( y=0 \)): \( x = \frac{2 x_1 y_1}{y_1} = 2 x_1 \).
- Y-intercept (put \( x=0 \)): \( y = \frac{2 x_1 y_1}{x_1} = 2 y_1 \).
5. Area of triangle \( = \frac{1}{2} |X \text{-intercept} \times Y \text{-intercept}| \).
\[ Area = \frac{1}{2} |2 x_1 \times 2 y_1| = 2 |x_1 y_1| \]
6. Since \( (x_1, y_1) \) is on the curve, \( x_1 y_1 = a^2 \).
\[ Area = 2a^2 \text{ sq. units.} \]
Step 4: Final Answer:
The area is \( 2a^2 \).