Question

The area of the triangle formed by the tangent to the curve \[ xy=a^2 \] at \((x_1,y_1)\) on it and the axes is

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

Hint:

For the rectangular hyperbola \[ xy=a^2, \] the tangent at \((x_1,y_1)\) is \[ xy_1+yx_1=2a^2. \] Intercept form of the tangent helps quickly find the area formed with coordinate axes.

Answer 1

The Correct Option is C

Step 1: Write the equation of the curve.
The curve is $xy = a^2$, so $y = a^2/x$. At point $(x_1, y_1)$ on the curve, $x_1 y_1 = a^2$.

Step 2: Find the slope of the tangent.
Differentiating $xy = a^2$ implicitly: $y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$. At $(x_1, y_1)$: slope $= -\frac{y_1}{x_1}$.

Step 3: Write the tangent line equation.
Tangent at $(x_1, y_1)$: $y - y_1 = -\dfrac{y_1}{x_1}(x - x_1)$. Rearranging: $xy_1 + yx_1 = 2x_1y_1 = 2a^2$.

Step 4: Find the x-intercept.
Set $y = 0$: $xy_1 = 2a^2 \Rightarrow x = \dfrac{2a^2}{y_1} = \dfrac{2x_1 y_1}{y_1} = 2x_1$.

Step 5: Find the y-intercept.
Set $x = 0$: $yx_1 = 2a^2 \Rightarrow y = \dfrac{2a^2}{x_1} = 2y_1$.

Step 6: Compute the area of the triangle.
The triangle has base $2x_1$ along x-axis and height $2y_1$ along y-axis. Area $= \dfrac{1}{2} \times 2x_1 \times 2y_1 = 2x_1 y_1 = 2a^2$.
\[ \boxed{2a^2 \text{ sq. units}} \]