Find the angle of intersection of the curves `2y^(2) = x^(3) and 32x`.

Explanation

`2y^(2)= x^(3) …(1)`
`y^(2) = 32x` …(2)
Solving eqs. (1) and (2)
` 64x = x^(3)`
` rArr x(x^(2)-64) = 0 rArr x=0,8,-8`
`:. "At "x=0,y-0`
`"At " x=8, y=pm 16`
At x =- 8, y is imaginary.
Therefore the points of intersection are (0, 0), (8, 16) and (8, -16).
From eq.(1) `" "` From eq.(2)
`3y^(2) = x^(3)" "y^(2) = 32x`
` rArr 4y (dy)/(dx) = 3x^(2) rArr 2y (dy)/(dx) = 32`
`rArr (dy)/(dx) = (3x^(2))/(4y) rArr (dy)/(dx) = (16)/y`
At point (8, 16)
` m_(1) = (3(8)^(2))/(4 xx 16) = 3 " "m_(2) = (16)/(16) = 1`
` and tan theta_(1) = (m_(1)-m_(2))/(1+m_(1)m_(2)) = (3-1)/(1+3) = 1/2 `
` rArr theta_(1) = tan^(-1) 1/2*`
At point (8,16)
`m_(3)= (3(8)^(2))/(4 xx (-16)) = - 3 " "m_(4) = 16/((-16))=-1`
`and tan theta_(2) = (m_(3)-m_(4))/(1+m_(3)m_(4)) = (-3+1)/(1+3) = (-1/2)`
`rArr theta_(2) = tan^(-1) (-1/2)*`
At point (0, 0), the slopes cannot be determined.