If the tangent at `(1,1)` on `y^2=x(2-x)^2` meets the curve again at `P ,` then find coordinates of `P`.

Options

A. `(4,4)`
B. `(-1,2)`
C. `(9//4,3//8)`
D. none of these

Answer

C

Explanation

We have,
` y^(2)=x(2-x)^(2) " "...(i) `
` rArr y^(2)=x^(3) - 4x^(2) +4x `
` rArr 2y(dy)/(dx)=3x^(2)-8x+4 `
` rArr (dy)/(dx)=(3x^(2)-8x+4)/(2y) rArr ((dy)/(dx))_((1","1)) =(3-8+4)/(2) = (-1)/(2) `
The equation of the tangent at (1,1) is
` y-1=(-1)/(2)(x-1) rArr x+2y-3=0 " " ...(ii) `
Solving (i) and (ii), we get `x=9//4 ` and `y=3//8.`
Hence, the coordinates of P are `(9//4,3//8). `