Question

Tangent to a curve intersect the $y-$axis at a point $P$. A line perpendicular this tangents through $P$ passes through another point $(1,0)$. then the differential equation of the curve is

• A. $y\frac{dy}{dx}-x{\left(\frac{dy}{dx}\right)}^2=0$
• B. $x\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=1$
• C. $y\frac{dy}{dx}-x=1$
• D. None of these

Answer 1

The correct option is A $y\frac{dy}{dx}-x{\left(\frac{dy}{dx}\right)}^2=0$

Let $R(x,f(x\left)\right)$ be the point at which tangent is drawn to the curve

The equation of tangent at $R$ is

$\Rightarrow Y-f\left(x\right)=f^{\prime }\left(x\right)(X-x)$

The co-ordinates of point $P$ are $(0,f(x)-x{f}^{\prime }(x\left)\right)$

The slope of perpendicular line through $P$ is

$\Rightarrow \frac{f\left(x\right)-x{f}^{\prime }\left(x\right)}{-1}=-\frac{1}{f^{\prime }\left(x\right)}$

$\Rightarrow f\left(x\right)f^{\prime }\left(x\right)-x\left(f^{\prime }\right(x){)}^2=1$

Therefore the required Differential equation is

$\Rightarrow y\frac{dy}{dx}-x(\frac{dy}{dx}{)}^2=1$