Find the point on the curve `9y^2=x^3,` where the normal to the curve makes equal intercepts on the axes.

Options

A. `(4, pm 8/3)`
B. `4,(-8)/3)`
C. `(4,pm3/8)`
D. `(pm 4, 3/8)`

Answer

A

Explanation

Given curve `9y^(2)= x^(3)` …(1)
` rArr 18y(dy)/(dx) = 3x^(2) rArr (dy)/(dx) = x^(2)/(6y)` …..(2)
Let ` ( alpha, beta)` be the point on equation (1) at which the normal cuts equal intercepts on the axes then
` 9 beta^(2)=alpha^(3)` ....(3)
From equation (2), slope of normal at ` (alpha, beta)`
`(-1)/((dy)/(dx))_(alpha, beta) = (-1)/(alpha^(2)//(6beta)) = (-6beta)/alpha^(2) ` ....(4)
` :' ` the normal to the curve cuts equal intercepts on the axes.
` :. " slope of normal "= tan 45^(@) or tan 135^(@) = pm 1 `...(5)
From equation (4), ` (-6 beta)/alpha^(2) = pm 1`
` rArr beta = pm alpha^(2)/6`
Put the value of `beta` in equation(3)
` 9(pm alpha^(2)/6)^(2) = alpha^(3)`
` rArr alpha^(4) = 4alpha^(3) rArr alpha^(3) (alpha-4) = 0`
` rArr alpha=0 or alpha = 4`
When ` alpha = 0, beta=0` , then the normal passes through (0, 0), . It means they do not intersect anywhere.
Put ` alpha= 4`,
` beta = pm 4^(2)/6 = pm 8/3` .