Q.
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
The equation of the tangent at the point R(x,f(x)) is Y−f(x)=f′(x)(X−x)
The coordinates of the point P are (0,f(x)−xf′(x))
The slope of the perpendicular line
through P is −1f(x)−xf′(x)=−f′x1 ⇒f(x)f′(x)−x(f′(x))2=1 ⇒dxydy−x(dxdy)2=1
which is required differential equation to the curve at y=f(x)