At what point on the curve `x^(3) - 8a^(2) y= 0` the slope of the normal is `-2//3` ?

Options

A. (a,a)
B. (2a,-a)
C. (2a,a)
D. None of these

Answer

C

Explanation

Given curve is `x^(2) - 8a^(2)y = 0`
On differentiating w.r.t . X, we get
` 3x^(2) - 8a^(2)(dy)/(dx) = 0 rArr (dy)/(dx) = (3x^(2))/(8a^(2))`
`therefore` Slope of the normal
` =-(1)/(((dy)/(dx))) = -(1)/((3x^(2))/(8a^(2))) = - (8a^(2))/(3x^(2))`
Given ` - (8a^(2))/(3x^(2)) = - (2)/(3)`
`rArr " "x^(2) = 4a^(2) rArr x = pm 2a`
At `" "x = pm 2a, " "y = pm a`
`therefore " "(x,y) = (2a,a)`