# find the equation of the tangent to the parabola x=2at, y=at^2 the point where t=3

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convert equation in cartesian form ,we have

`x=2at ,y=at^2`

`x^2=4a^2t^2`

`t^2=x^2/(4a^2)` ,substitute in y

we have

`y=a(x^2/(4a^2))`

`x^2=4ay` (i)

differentiate (i) with respect to x,

`4a(dy)/(dx)=2x`

`(dy)/(dx)=x/(2a)`

`(dy)/(dx)}_ t=(2at)/(2a)=t`

Equation of the tangents at 't'

`y-at^2=t(x-2at)`

`y=tx-2at^2+at^2`

`y=tx-at^2`

Ans.