Find the equations of the tangent and normal to the parabola `y^2=4a x`at the point `(a t^2,2a t)`.

Explanation

Equation of parabola: ` y^(2) = 4ax`
` rArr 2y(dy)/(dx) = 4a rArr (dy)/(dx) = (2a)/y`
at point `(at^(2), 2at)`
Slope of tangent = `(2a)/(2at) = 1/t`
and equation of tangent
` y-2at = 1/t (x-at^(2))`
`rArr ty - 2at^(2) = x - at^(2)`
` rArr x- ty + at^(2) = 0`
Slope of normal = `(-1)/(1//t) = - t`
Equation of normal
` y - 2at = - t(x-at^(2))`
` rArr y- 2at = - tx + at^(3)`
` rArr tx + y - 2at + at^(3) = 0`