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

Options

A. `(1)/(t)`
B. t
C. `-t`
D. `-(1)/(t)`

Answer

C

Explanation

Equation of parabola is
`y^(2)=4ax`
`2y(dy)/(dx)=4a`
`therefore (dy)/(dx)=(2a)/(y)`
so slope of normal `=-((dx)/(dy))_(at^(2),2at)`
`=-((y)/(2a))=-(2at)/(2a)=-t`