Prove that the equation of the normal to `x^(2/3)+y^(2/3)=a^(2/3)` is `ycostheta-xsintheta=acos2theta,` where `theta` is the angle which the normal makes with the axis of `xdot`

Explanation

` x^(2//3)+y^(2//3)=a^(2//3)`…(1)
` rArr 2/3* x^(-1//3) + 2/3 * y^(-1//3)(dy)/(dx) = 0 `
` rArr (dy)/(dx) =- (y^(1//3)/x^(1//3))`
Thus the slope of normal = ` (x^(1//3))/(y^(1//3)) = tan phi ` (Given)
`rArr x/y = tan^(3) phi`
` rArr x= y tan^(3) phi`
From eq. (1)
`y^(2//3)*tan^(2) phi + y^(2//3)=a^(2//3)`
`rArr y^(2//3)*sec^(2) phi = a^(2//3)`
`rArr y= a cos^(3) phi`
and ` x = a cos ^(3) phi * tan ^(3) phi = a sin^(3) phi`
Now equation of normal at point` (a sin^(3) phi, cos^(3) phi)` is
` y-a cos^(3) phi = tan phi ( x-a sin ^(3) phi )`
`rArr y cos phi -a cos^(4) phi = x sin phi -a sin ^(4) phi`
`rArr y cos phi -x sin phi = a (cos^(4) phi - sin ^(4) phi)`
`= a (cos^(2) phi - sin^(2) phi ) * (cos^(2) phi + sin ^(2) phi)`
`= a cos 2 phi`.