Prove that the equation of a line passes through the point `(a cos^(3)alpha`, `a sin^(3)alpha`) and perpendicular to the line `x tan alpha+y=a sin alpha` is `x cos alpha-y sin alpha-a cos 2 alpha`.

Explanation

Given line is
`x tan alpha+y=a sin alpha`
`implies y=-x tan alpha+asin alpha`
Its slope `m_(1)=-tan alpha`
Slope of perpendicular line `m_(2)=-(1)/(m_(1))`
`=-(1)/(-tanalpha)`
`=cot alpha`
Equation of required line
`y-a sin^(3)alpha=cot alpha(x-a cos^(3)alpha)`
`impliesy-a sin^(3)alpha=(cosalpha)/(sin alpha)(x-acos^(3)alpha)`
`implies ysin alpha-a sin^(4)alpha=xcosalpha-acos^(4)alpha)`
`implies x cos alpha-y sin alpha=a(cos^(4)alpha-sin^(4)alpha)`
`=a(cos^(2)alpha-sin^(2)alpha)(cos^(2)alpha+sin^(2)alpha)`
`=acos2alpha`