Question

Tangent to a curve $y=f\left(x\right)$ intersects the $y-$axis at a point $P.$ A line perpendicular to this tangent through $P$ passes through another point $(1,0).$ The differential equation of the curve is

• A. $y\frac{dy}{dx}-x{\left(\frac{dy}{dx}\right)}^2=1$
• B. $\frac{x{d}^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$
• C. $y\frac{dy}{dx}+x=1$
• D. $\frac{xdy}{dx}+{\left(\frac{dy}{dx}\right)}^2=1$

Answer 1

The correct option is A $y\frac{dy}{dx}-x{\left(\frac{dy}{dx}\right)}^2=1$

The equation of the tangent at the point
$R(x,f\left(x\right))$ is $Y-f\left(x\right)=f^{\prime }\left(x\right)(X-x)$
The coordinates of the point $P$ are $(0,f\left(x\right)-x{f}^{\prime }\left(x\right))$
The slope of the perpendicular line through $P$ is $\frac{f\left(x\right)-x{f}^{\prime }\left(x\right)-0}{0-1}=-\frac{1}{f^{\prime }\left(x\right)}$
$\Rightarrow f\left(x\right)f^{\prime }\left(x\right)-x{\left(f^{\prime }\left(x\right)\right)}^2=1$
$\Rightarrow \frac{ydy}{dx}-x{\left(\frac{dy}{dx}\right)}^2=1$ which is the required differential equation of the curve $y=f\left(x\right)$