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 tangent.
Given curve: \[ xy=a^2 \] At the point \[ (x_1,y_1), \] we have \[ x_1y_1=a^2 \] The equation of tangent to \[ xy=a^2 \] at \((x_1,y_1)\) is \[ xy_1+yx_1=2a^2 \]

Step 2: Find the intercepts on the axes.
To find the \(x\)-intercept, put \[ y=0 \] Then, \[ xy_1=2a^2 \] \[ x=\frac{2a^2}{y_1} \] Since \[ a^2=x_1y_1, \] we get \[ x=\frac{2x_1y_1}{y_1} \] \[ x=2x_1 \] Thus, the tangent cuts the \(x\)-axis at \[ (2x_1,0) \] Now, to find the \(y\)-intercept, put \[ x=0 \] Then, \[ yx_1=2a^2 \] \[ y=\frac{2a^2}{x_1} \] Using \[ a^2=x_1y_1, \] we get \[ y=\frac{2x_1y_1}{x_1} \] \[ y=2y_1 \] Thus, the tangent cuts the \(y\)-axis at \[ (0,2y_1) \]

Step 3: Find the area of the triangle.
Area of triangle formed with coordinate axes is \[ \frac{1}{2}\times (\text{\(x\)-intercept}) \times (\text{\(y\)-intercept}) \] Therefore, \[ \text{Area} = \frac{1}{2}\times (2x_1)\times (2y_1) \] \[ =2x_1y_1 \] Since \[ x_1y_1=a^2, \] we get \[ \text{Area}=2a^2 \]

Step 4: Final conclusion.
Therefore, \[ \boxed{2a^2} \] sq. units