Question

Let the curve $y=f\left(x\right)$ pass through the origin. If the mid point of the line segment of its normal between any point on the curve and the $x-$axis lies on the parabola $y^2=4x$, then the equation of the curve $y=f\left(x\right)$ satisfies

• A. $y^2=16x+64-64e^{x/4}$
• B. $y^2=16x+64+64e^{x/4}$
• C. $y^2=16x-64-64e^{x/4}$
• D. $y^2=-16x-64+64e^{x/4}$

Answer 1

The correct option is A $y^2=16x+64-64e^{x/4}$

Let $(x,y)$ be a point on the curve $y=f\left(x\right)$ and $(a,0)$ be a point on the $x-$axis.
Slope of the normal is
$-\frac{dx}{dy}=\frac{y-0}{x-a}\Rightarrow x-a=-y\frac{dy}{dx}\Rightarrow a=x+y\frac{dy}{dx}\cdots \left(1\right)$

Now, the mid-point of the line segment of the normal is $(\frac{x+a}{2},\frac{y}{2})$
Putting the mid-point in the given equation of the parabola,
$\frac{y^2}{4}=4\left(\frac{x+a}{2}\right)\Rightarrow a=\frac{y^2}{8}-x\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$,
$x+y\frac{dy}{dx}=\frac{y^2}{8}-x\Rightarrow 2y\frac{dy}{dx}-\frac{y^2}{4}=-4x$

Assuming $y^2=t\Rightarrow 2y\frac{dy}{dx}=\frac{dt}{dx}$
So, the differential equation becomes
$\frac{dt}{dx}-\frac{t}{4}=-4x$
$I.F.=\exp \left(\int -\frac{1}{4}dx\right)=e^{-x/4}$
So, the solution of the differential equation is,
$t\cdot e^{-x/4}=\int -4x{e}^{-x/4}dx+C$
where $C$ is a constant of integration.
$\Rightarrow t\cdot e^{-x/4}=-4[-4x{e}^{-x/4}+4\int e^{-x/4}dx]+C\Rightarrow y^2\cdot e^{-x/4}=-4[-4x{e}^{-x/4}-16e^{-x/4}]+C$
The curve passes through the origin. So putting $(0,0)$, we get
$C=-64$
Therefore, $y^2\cdot e^{-x/4}=-4[-4x{e}^{-x/4}-16e^{-x/4}]-64$
$\Rightarrow y^2=16x+64-64e^{x/4}$