Question

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

• A. $(2,2)$
• B. $(-1,-2)$
• C. $\left(\frac{9}{4}, \frac{3}{8}\right)$
• D. None of these

Answer 1

Answer: (C)

$y^{2} = x\left(2 - x\right)^{2}$
$\Rightarrow y^{2} = x^{3}-4x^{2} + 4x \quad\cdots\left(1\right)$
$\Rightarrow 2y \frac{dy}{dx} = 3x^{2} - 8x + 4$
$\Rightarrow \frac{dy}{dx} = \frac{3x^{2}-8x+4}{2y}$
$\Rightarrow \left[\frac{dy}{dx}\right]_{p} = \frac{3-8+4}{2} = \frac{1}{2}$
$\therefore $ Equation of tangent at $P$ is
$y = 1 = -\frac{1}{2}\left(x-1\right)$
$\Rightarrow x + 2y - 3 = 0 \quad\cdots \left(2\right)$
Using $y = \frac{3-x}{2}$ in $\left(1\right)$, we get
$\left(\frac{3-x}{2}\right)^{2} = x^{2}- 4x^{2} + 4x$
$\Rightarrow 9 + x^{2} - 6 x = 4x^{3} - 16x^{2} + 16x$
$\Rightarrow 4x^{3} - 17x^{2} + 22x - 9 = 0$
which has two roots $1, 1$ (Because of $\left(2\right)$ being tangentat $\left(1,1\right)$).
Sum of $3$ roots $= \frac{17}{4}$
$\Rightarrow 3^{rd}$ root $= \frac{17}{4} - 2 = \frac{9}{4}$
Then, $y = \frac{3-\frac{9}{4}}{2} = \frac{3}{8}$
$\therefore Q$ is $\left(\frac{9}{4}, \frac{3}{8}\right)$.