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. (2, 0)
C. `(9/4, 3/8)`
D. `(3,3^(1//2))`

Answer

C

Explanation

`2y(dy)/(dx)=(2-x)^(2)-2(2-x)`
`"So "(dy)/(dx):|_("(1,1)")``=-(1)/(2)`
Therefore, the equation of tangent at (1,1) is
`y-1=-(1)/(2)(x-1)`
`rArr" "y=(-x+3)/(2)`
The intersection of the tangent and the curve is given by
`(1//4)(-x+3)^(2)=x(4+x^(2)-4x)`
`rArr" "4x^(3)-17x^(2)+22x-9=0`
`rArr" "(x-1)(4x^(2)-13x+9)=0`
`rArr" "(x-1)(4x^(2)-13x+9)=0`
`rArr" "(x-1)^(2)(4x-9)=0`
Since x = 1 is already the point of tangency, `x=9//4` and `y^(2)=(9)/(2)(2-(9)/(4))^(2)=(9)/(64)`.
Thus, the required point is `(9//4, 3//8)`.