Question

Tangent to a non-linear curve $y=f\left(x\right)$ at any point $P$ intersects $x-$axis and $y-$axis at $A$ and $B$ respectively. If normal to the curve $y=f\left(x\right)$ at $P$ intersects $y-$axis at $C$ such that $AC=BC$ and $f\left(2\right)=3,$ then the equation of curve is

• A. $xy=6$
• B. $x^2+y^2=13$
• C. $2y^2=9x$
• D. $2y=3x$

Answer 1

The correct option is A $xy=6$


Equation of tangent at $P(x_1,y_1)$ is
$y-y_1=\frac{dy}{dx}(x-x_1)$
$\Rightarrow A(x_1-\frac{y_1}{\left(\frac{dy}{dx}\right)},0)$ and $B(0,y_1-x_1\frac{dy}{dx})$

Equation of normal at $P(x_1,y_1)$ is
$y-y_1=\frac{-1}{\frac{dy}{dx}}(x-x_1)$
$\Rightarrow C(0,y_1+\frac{x_1}{\frac{dy}{dx}})$

Now, $AC=BC$
$\Rightarrow {(x_1-\frac{y_1}{\frac{dy}{dx}})}^2+{(y_1+\frac{x_1}{\frac{dy}{dx}})}^2={(x_1\frac{dy}{dx}+\frac{x_1}{\frac{dy}{dx}})}^2$
$\Rightarrow y_1^2=x_1^2\frac{dy}{dx}$
$\Rightarrow x_1\frac{dy}{dx}=\pm y_1$
For non-linear curve, we take negative sign.
$x\frac{dy}{dx}=-y$
$\Rightarrow \frac{dx}{x}+\frac{dy}{y}=0$
$\Rightarrow \ln \left(xy\right)=\ln C$
$\Rightarrow xy=C$
At $x=2$ and $y=3,$ we get $C=6$
$\therefore xy=6$