It is given that the equation of a curve $C$ is $4{{x}^{2}}-8{{x}^{2}}y-32=5{{y}^{3}}$. Show tha

Question

It is given that the equation of a curve $C$ is $4{{x}^{2}}-8{{x}^{2}}y-32=5{{y}^{3}}$. Show that there is no point on $C$ where the tangent is parallel to the $y$-axis.[4] Find the equation of the tangent which is parallel to the $x$-axis.[3] The point $P\,(x,y)$ moves along $C$ in a way such that the $x$- coordinate of $P$ is increasing at a constant rate of $2$ units per second. Find the exact rate of increase of the $y$-coordinate at the instant when $x=1.5$.[3]

Accepted Answer

(i) No vertical tangent exists since $15y^2+8x^2=0$ only at $(0,0)$, which is not on $C$ (ii) $y=\sqrt[3]{-\dfrac{32}{5}}$ (iii) $\dfrac{24}{11}\text{ units s}^{-1}$

2021 TJC P1 Q7

2021 TJC P1 Q7

Video Solution:

Solution:

Solution locked