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)

$y^2=x{\left(2-x\right)}^2$
$\Rightarrow y^2=x^3-4x^2+4x\cdots \left(1\right)$
$\Rightarrow 2y\frac{dy}{dx}=3x^2-8x+4$
$\Rightarrow \frac{dy}{dx}=\frac{3x^2-8x+4}{2y}$
$\Rightarrow {\left[\frac{dy}{dx}\right]}_p=\frac{3-8+4}{2}=\frac{1}{2}$
$\therefore$ Equation of tangent at $P$ is
$y=1=-\frac{1}{2}\left(x-1\right)$
$\Rightarrow x+2y-3=0\cdots \left(2\right)$
Using $y=\frac{3-x}{2}$ in $\left(1\right)$, we get
${\left(\frac{3-x}{2}\right)}^2=x^2-4x^2+4x$
$\Rightarrow 9+x^2-6x=4x^3-16x^2+16x$
$\Rightarrow 4x^3-17x^2+22x-9=0$
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 $\left(\frac{9}{4},\frac{3}{8}\right)$.