The curve $C$ has equation ${{\tan }^{-1}}y=x\ln {{x}^{2}}-2x-y$, where $x\in \mathbb{R}$

Question

The curve $C$ has equation ${{\tan }^{-1}}y=x\ln {{x}^{2}}-2x-y$, where $x\in \mathbb{R}$, $x>0$. Show that $\left( 2+{{y}^{2}} \right)\frac{\mathrm{d}y}{\mathrm{d}x}=2\left( 1+{{y}^{2}} \right)\ln x$.[2] Hence find the coordinates of the stationary point of $C$.[3] By differentiating the result shown in part (i), determine the nature of the stationary point.[3] State the equation of the normal to the curve $C$ that is parallel to the $y$-axis.[1]

Accepted Answer

(ii) Coordinates of stationary point: ${(1,-1.15)}$ (iii) ${\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 2\left(\frac{1+y^2}{2+y^2}ight) > 0 \Rightarrow 	ext{Minimum Point}}$ (iv) ${x = 1}$

2020 EJC Promo Q7

2020 EJC Promo Q7

Video Solution:

Solution:

Solution locked