Question

The area of the triangle formed by the coordinate axes and a tangent to the curve $\mathrm{xy}={\mathrm{a}}^2$ at the point $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ on it is

• A. $\frac{{\mathrm{a}}^2{\mathrm{x}}_1}{{\mathrm{y}}_1}$
• B. $\frac{{\mathrm{a}}^2{\mathrm{y}}_1}{{\mathrm{x}}_1}$
• C. $2{\mathrm{a}}^2$
• D. $4{\mathrm{a}}^2$

Answer 1

The correct option is C

$2{\mathrm{a}}^2$

Explanation for correct option

Finding the area of the triangle

Given, $\mathrm{y}=\frac{{\mathrm{a}}^2}{\mathrm{x}}$

Differentiating the above equation $w.r.t.x$:

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}$

At $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$,$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{{\mathrm{a}}^2}{{{\mathrm{x}}_1}^2}$

Thus, the tangent to the curve will be,

$$\begin{gathered} y-y_1=\left(\frac{-a^2}{x_1^2}\right)(x-x_1) \\ \Rightarrow y\times x_1^2-y_1{x}_1^2=-a^{2}x+a^2{x}_1 \\ \Rightarrow a^{2}x+x_1^{2}y=x_1\left[a^2+x_1{y}_1\right] \end{gathered}$$

Point on x-axis is $\left({2\mathrm{x}}_1,0\right)$ and y-axis is $\left(0,\frac{2a^2}{x_1}\right)$.

Therefore, the required area of the triangle will be:

$$\begin{gathered} \mathrm{Area}=\frac{1}{2}\times {2\mathrm{x}}_1\times \frac{{2\mathrm{a}}^2}{{\mathrm{x}}_1} \\ ={2\mathrm{a}}^2 \end{gathered}$$

Hence, the correct answer is Option (C).