The area of the triangle formed by the co-ordinate axes and a tangent to the curve $x \mathrm{y}=a^2$ at the point $\left(x_1, \mathrm{y}_1\right)$ is ________ sq.units (where $a, x_1$ and $y_1$ are non zero)

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Aicharya · Accepted Answer

The tangent to the curve $xy = a^2$ at point $(x_1, y_1)$ has equation $y_1x + x_1y = 2a^2$, derived from differentiating $xy = a^2$ to obtain $\frac{dy}{dx} = -\frac{y}{x}$ and using point-slope form.

The x-intercept occurs at $\left(\frac{2a^2}{y_1}, 0\right)$ and the y-intercept at $\left(0, \frac{2a^2}{x_1}\right)$, found by setting $y = 0$ and $x = 0$ respectively in the tangent equation.

The area of the triangle formed by the coordinate axes and this tangent is $\frac{1}{2} \cdot \left|\frac{2a^2}{y_1}\right| \cdot \left|\frac{2a^2}{x_1}\right| = \frac{4a^4}{2|x_1y_1|}$.

Since $x_1y_1 = a^2 > 0$, the area simplifies to $\frac{2a^4}{a^2} = 2a^2$.

Final answer: $2a^2$

The area of the triangle formed by the co-ordinate axes and a tangent to the curve $x \mathrm{y}=a^2$ at the…

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