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: Differentiate the Curve
$xy = a^2 \implies y + x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x}$.
At $(x_1, y_1)$, slope $m = -\frac{y_1}{x_1}$.

Step 2: Equation of Tangent
$y - y_1 = -\frac{y_1}{x_1}(x - x_1)$
$x_1y - x_1y_1 = -xy_1 + x_1y_1 \implies xy_1 + yx_1 = 2x_1y_1$.

Step 3: Find Intercepts
X-intercept ($y=0$): $x = 2x_1$.
Y-intercept ($x=0$): $y = 2y_1$.

Step 4: Calculation
Area $= \frac{1}{2} \times | \text{Base} \times \text{Height} | = \frac{1}{2} | (2x_1)(2y_1) | = 2 | x_1y_1 |$.
Since $(x_1, y_1)$ is on the curve $xy = a^2$, then $x_1y_1 = a^2$.
Area $= 2a^2$.
Final Answer: (C)