Question

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

• A. $\frac{a^2{x}_1}{y_1}$
• B. $\frac{a^2{y}_1}{x_1}$
• C. $2a^2$
• D. $4a^2$

Answer 1

The correct option is C $2a^2$

Since $y=\frac{a^2}{x},\therefore \frac{dy}{dx}=-\frac{a^2}{x^2}$
$\therefore$ At $(x_1,y_1),\frac{dy}{dx}=\frac{-a^2}{x_1^2}$
Thus tangent to the curve will be $y-y_1=\frac{-a^2}{x_1^2}(x-x_1)$
$\Rightarrow y{x}_1^2-y_1{x}_1^2=a^{2}x+a^2{x}_1$
$\Rightarrow a^{2}x+x_1^{2}y=x_1(x_1{y}_1+a^2)=2a^2{x}_1,(\because x_1{y}_1=a^2)$
This meets the x-axis where y=0
$\therefore a^{2}x=2a^2{x}_1,\therefore x=2x_1$
$\therefore$ Point on the x-axis is $(2x_1,0)$
Again tangent meets the y-axis where x =0
$\because x_1^2=2a^2{x}_1,\therefore y=\frac{2a^2}{x_1}$
So point on the y-axis is $(0,\frac{2a^2}{x_1})$
Required area $=\frac{1}{2}\left(2x_1\right)\left(\frac{2a^2}{x_1}\right)=2a^2$