Mathematics - Application of Derivative Question with Solution | TestHub

$$\begin{array}{rl}& y^2=x(2-x{)}^2\Rightarrow y^2=x^3-4x^2+4x\\ & \Rightarrow 2y\frac{dy}{dx}=3x^2-8x+4\Rightarrow \frac{dy}{dx}=\frac{3x^2-8x+4}{2y}\\ & \Rightarrow {\left[\frac{dy}{dx}\right]}_P=\frac{3-8+4}{2}=-\frac{1}{2}\end{array}$$ $\therefore$ 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: ${\left(\frac{3-x}{2}\right)}^2=x^3-4x^2+4x$ $$\Rightarrow 4x^3-17x^2+22x-9=0$$ 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 $=\frac{17}{4}-2=\frac{9}{4}$ $$\text{ Then, }\mathrm{y}=\frac{3-\frac{9}{4}}{2}=\frac{3}{8}\therefore \mathrm{Q}\text{ is }\left(\frac{9}{4},\frac{3}{8}\right)$$