Question

Given the curve ${\left(\frac{x^2-y^2}{xy}\right)}^3=\frac{512}{27},$ then

• A. the slope of normal drawn to the curve at $(3,1)$ is $\frac{1}{3}$
• B. the value of
  ${(2\frac{dy}{dx}-5\frac{d^{2}y}{d{x}^2})}_{(3,1)}=\frac{2}{3}$
• C. the slope of normal drawn to the curve at $(3,1)$ is $-3$
• D. the equation of tangent at $(3,1)$ to given curve $3y+x=0$.

Answer 1

Answer: (B), (C)

the slope of normal drawn to the curve at $(3,1)$ is $-3$

Given equation is homogeneous
Putting $y=vx$ in the given equation and differentiating both sides, we get

$\frac{dy}{dx}=\frac{y}{x}$ and $\frac{d^{2}y}{d{x}^2}=0$
${\left(\begin{array}{c}\frac{dy}{dx}\end{array}\right)}_{(3,1)}=\frac{1}{3}$
$\therefore {(2\frac{dy}{dx}-5\frac{d^{2}y}{d{x}^2})}_{(3,1)}=\frac{2}{3}$

Slope of normal is $-3$