At each point (x,y)of a curve, the intercept of the tangent on they-axis is equal to 2xy2. Find the equation of the curve.
xy=x2+c
yx=y2+c
xy=y2+c
yx=x2+c
Given that y-intercept , c=2xy2
As the Equation of the tangent is y=mx+c
and Slope m=dydx
So we have,
y=dydxx+2xy2⇒ydx-xdy=2xy2dx⇒(ydx-xdy)y2=2xdx⇒∫(ydx-xdy)y2=∫2xdx[integratingbothside]⇒∫dxy=∫2xdx⇒xy=x2+c
Hence, option (a) is the correct option.
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