Q29 / 75 · VITEEE 2021 · Mathematics
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$(-1,-2)$
- Correct answerC$\left(\frac{9}{4}, \frac{3}{8}\right)$
- DNone of these
Solution
$$
\begin{aligned}
& \text { } y^2=x(2-x)^2 \Rightarrow y^2=x^3-4 x^2+4 x \\
& \Rightarrow 2 y \frac{d y}{d x}=3 x^2-8 x+4 \Rightarrow \frac{d y}{d x}=\frac{3 x^2-8 x+4}{2 y} \\
& \Rightarrow\left[\frac{d y}{d x}\right]_P=\frac{3-8+4}{2}=-\frac{1}{2}
\end{aligned}
$$
$\therefore$ Equation of tangent at $\mathrm{P}$ is: $\mathrm{y}-1=-\frac{1}{2}(\mathrm{x}-1)$
$$
\Rightarrow \mathrm{x}+2 \mathrm{y}-3=0
$$
Using $y=\frac{3-x}{2}$ in (i),
we get: $\left(\frac{3-x}{2}\right)^2=x^3-4 x^2+4 x$
$$
\Rightarrow 4 x^3-17 x^2+22 x-9=0
$$
which has two roots 1,1
(Because of (ii) being tangent at $(1,1)$ ).
Sum of 3 roots $=\frac{17}{4} \therefore 3$ rd root $=\frac{17}{4}-2=\frac{9}{4}$
$$
\text { Then, } \mathrm{y}=\frac{3-\frac{9}{4}}{2}=\frac{3}{8} \therefore \mathrm{Q} \text { is }\left(\frac{9}{4}, \frac{3}{8}\right)
$$
Hence, the correct answer is option: 'C'.