Question

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

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

Answer 1

Answer: (A)

Equation of tangent at $P\left(x_1,y_1\right)$
$y-y_1=\frac{dy}{dx}\left(x-x_1\right)$

$\Rightarrow A\left(\left(x_1-\frac{y_1}{\left(\frac{dy}{dx}\right)}\right),0\right)$
$B\left(0,y_1-x_1\frac{dy}{dx}\right)$
Equation of normal at $P\left(x_1,y_1\right)$
$y-y_1=\frac{-1}{dy/dx}\left(x-x_1\right)$
$C\left(0,\frac{x_1}{dy/dx}+y_1\right)$
$\Theta AC=BC$
${\left(x_1-\frac{y_1}{dy/dx}\right)}^2+{\left(\frac{x_1}{dy/dx}+y_1\right)}^2={\left(y_1-\frac{dy}{dx}{x}_1-\frac{x_1}{dy/dx}-y_1\right)}^2$
$y_1=\pm x_1\frac{dy}{dx}$
$\Rightarrow \int \frac{-dy}{y}=\int \frac{dx}{x}\text{ (positive sign give linear curve) }$
$\Rightarrow -\ln y=\ln x+\ln c\Rightarrow xy=c$
$\therefore f(2)=3\Rightarrow xy=6$