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. (-1,2)
• C. $(\frac{9}{4},\frac{3}{8})$
• D. (0,0)

Answer 1

The correct option is C $(\frac{9}{4},\frac{3}{8})$

2y $\frac{dy}{dx}=(2-x{)}^2$ - 2x(2 - x), so $\frac{dy}{dx}{|}_{(1,1)}=\frac{1}{2}[1-2]=-\frac{1}{2}$
Therefore, the equation of tangent at (1, 1) is y - 1 = -$\frac{1}{2}(x-1)$
$\Rightarrow$ 2y - 2 = -x + 1
$\Rightarrow$ y = $\frac{-x+3}{2}$
The intersection of the tangent and the curve is given by $\left(\frac{1}{4}\right)(-x+3{)}^2=x(4+x^2-4x)$
$\Rightarrow x^2-6x+9=16x+4x^3-16x^2$
$\Rightarrow 4x^3-17x^2+22x-9=0$
$\Rightarrow (x-1)(4x^2-13x+9)=0$
$\Rightarrow (x-1{)}^2(4x-9)=0$
Since x = 1 is already the point of tangency x = $\frac{9}{4}$ and $y^2$ = $\frac{9}{4}{(2-\frac{9}{4})}^2=\frac{9}{64}$
Thus the required point is $(\frac{9}{4},\frac{3}{8})$