Question

A normal is drawn at a point $P$ on a curve $y=f\left(x\right)$, meeting the $x-$axis and the $y-$axis at points $A$ and $B$ respectively. Let $\frac{1}{OA}+\frac{1}{OB}=1$, where $O$ is the origin. If the equation of the curve passes through $(2,3)$, then the number of points of intersection of $y=f\left(x\right)$ with $y-$axis is

Answer 1

Let point $P$ be $(x,y)$.
Then, equation of the normal at $P$ is,
$\frac{Y-y}{X-x}=-\frac{dx}{dy}$
$\Rightarrow (X-x)+(Y-y)\frac{dy}{dx}=0\Rightarrow X+Y\frac{dy}{dx}=x+y\frac{dy}{dx}$
Putting $Y=0$, coordinates of $A$ is $(x+y\frac{dy}{dx},0)$
Putting $X=0$, coordinates of $B$ is $(0,\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}})$

Given, $\frac{1}{OA}+\frac{1}{OB}=1$
$\Rightarrow \frac{1}{x+y\frac{dy}{dx}}+\frac{\frac{dy}{dx}}{x+y\frac{dy}{dx}}=1$

$\Rightarrow 1+\frac{dy}{dx}=x+y\frac{dy}{dx}\Rightarrow (y-1)\frac{dy}{dx}=1-x\Rightarrow (y-1)dy=(1-x)dx$
Integrating, we get
$\frac{y^2}{2}-y=x-\frac{x^2}{2}+C$
where $C$ is constant of integration.
Since the curve passes through $(2,3)$,
$\frac{9}{2}-3=2-\frac{4}{2}+C\Rightarrow C=\frac{3}{2}$
Therefore, the equation of the curve is,
$\frac{y^2}{2}-y=x-\frac{x^2}{2}+\frac{3}{2}$

Intersection with $y-$axis, putting $x=0$,
$\frac{y^2}{2}-y=\frac{3}{2}\Rightarrow y^2-2y-3=0\Rightarrow (y-3)(y+1)=0\Rightarrow y=-1,3$
Hence, the number of points of intersection with $y-$axis is $2$.