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
ydydx−x(dydx)2=1
The equation of the tangent at the point
R(h,f(h)) is y−f(h)=f′(h)(x−h)
The coordinates of the point P are (0,f(h)−hf′(h))
The slope of the perpendicular line is −f(h)+hf′(h)
Applying the condition for perpendicularity
⇒f′(h)f(h)−h(f′(h))2=1⇒ydydx−x(dydx)2=1
which is the required differential equation to the curve y=f(x)