Question

If the tangent at (1, 1) on $y^2=x(2-x{)}^2$ meets the curve again at P, then ‘P’ is

• A. $(4,4)$
• B. $(2,0)$
• C. $(\frac{9}{4},\frac{3}{8})$
• D. $(3,\sqrt{3})$

Answer 1

The correct option is C

$(\frac{9}{4},\frac{3}{8})$

$2y\left(\frac{dy}{dx}\right)=(2-x{)}^2-2x(2-x)$

$\Rightarrow \frac{dy}{dx}{|}_{(1,1)}=-\frac{1}{2}$

$\therefore$ Equation of tangent at (1, 1) is

$y=\frac{-x+3}{2}$

The intersection of the tangent and the curve is given by

$\frac{1}{4}(-x+3{)}^2=x(4+x^2-4x)$

$\Rightarrow (x-1)(4x^2-13x+9)=0$

$\Rightarrow (x-1{)}^2(4x-9)=0$

Since, $x=1$ is already the point of tangency, $x=\frac{9}{4}$ and $y=\frac{-x+3}{2}=\frac{3}{8}$

Thus, required point is $(\frac{9}{4},\frac{3}{8})$