Mathematics - Application of Derivative Question with Solution | TestHub

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