Q76 / 125 · BITSAT 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 answer
    C
    $\left(\frac{9}{4}, \frac{3}{8}\right)$
  • D
    None of these

Solution

$\begin{array}{l} y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x \ldots \text { (i) } \\ \Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}=3 x^{2}-8 x+4 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 x^{2}-8 x+4}{2 y} \\ \Rightarrow\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{P}}=\frac{3-8+4}{2}=-\frac{1}{2} \end{array}$ $y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x \ldots$ (i) $\Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}=3 x^{2}-8 x+4$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 x^{2}-8 x+4}{2 y}$ $\Rightarrow\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{P}}=\frac{3-8+4}{2}=-\frac{1}{2}$ $\therefore$ Equation of tangent at $\mathrm{P}$ is: $y-1=-\frac{1}{2}(x-1)$ $\Rightarrow x+2 y-3=0$ $\Rightarrow x+2 y-3=0$ Using $y=\frac{3-x}{2}$ in (i), we get: $\left(\frac{3-x}{2}\right)^{2}$ $\begin{array}{l} =x^{3}-4 x^{2}+4 x \\ \Rightarrow 4 x^{3}-17 x^{2}+22 x-9=0 \text {...(ii) } \end{array}$ which has two roots 1,1 (Because of (ii) being tangent at $(1,1)$ ). Sum of 3 roots $=\frac{17}{4}$ $\begin{array}{l} \therefore 3 \mathrm{rd} \text { root }=\frac{17}{4}-2=\frac{9}{4} \\ \text { Then, } y=\frac{3-\frac{9}{4}}{2}=\frac{3}{8} \\ \therefore \mathrm{Q} \text { is }\left(\frac{9}{4}, \frac{3}{8}\right) \end{array}$

Hence, the correct answer is option: 'C'.

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