Question

Tangent to a curve intercepts 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. $y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$
• C. $y\frac{dx}{dy}+x=1$
• D. None of these

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
$R(h,f(h\left)\right)$ is $y-f\left(h\right)=f^{\prime }\left(h\right)(x-h)$
The coordinates of the point $P$ are $(0,f(h)-h{f}^{\prime }(h\left)\right)$
The slope of the perpendicular line is $-f\left(h\right)+h{f}^{\prime }\left(h\right)$
Applying the condition for perpendicularity
$\Rightarrow f^{\prime }\left(h\right)f\left(h\right)-h\left(f^{\prime }\right(h){)}^2=1\Rightarrow \frac{ydy}{dx}-x{\left(\frac{dy}{dx}\right)}^2=1$
which is the required differential equation to the curve $y=f\left(x\right)$