At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point `( 4, 3)`. Find the equation of the curve given that it passes through `(2, 1)`.

Explanation

Slope of tangent `=(dy)/(dx)`
Slope of straight line joining the touch point `(x,y)` and the point `(-4,-3)=(y+3)/(x+4)`
Given, `(dy)/(dx)=(2(y+3))/(x+4)`
`implies (dy)/(y+3)=(2dx)/(x+4)`
`implies int(1)/(y+3)dy=2int(1)/(x+4)dx+c`
`implies log(y+3)=2log(x+4)+c`
It passes through the point `(-2,1)`.
`log4=2log2+c`
`implies c=0`
Therefore, `log(y+3)=2log(x+4)`
`implies y+3=(x+4)^(2)`
`implies y=x^(2)+8x+13`