The equation of the given parabola is,
y 2 =4ax
Equation of the slope of the line is given by,
slope= dy dx
Hence, slope of the parabola is determined by differentiating with respect to x.
2y( dy dx )=4a dy dx = 2a y
Slope of the tangent at point ( a t 2 ,2at ) is given by,
( dy dx ) ( a t 2 ,2at ) = 2a 2at = 1 t
Thus, the equation of the tangent at ( a t 2 ,2at ) with slope 1 t is,
y−at= 1 t ( x−a t 2 ) ty=x+a t 2
Slope of the normal at ( a t 2 ,2at ) is given by,
−1 slope of tangent at ( a t 2 ,2at ) =−t
Hence, the equation of normal at ( a t 2 ,2at ) with slope −t is given by,
y−2at=−t( x−a t 2 ) y=−tx+2at+a t 2
Thus, the equation of the tangent is ty=x+a t 2 and equation of the normal is y=−tx+2at+a t 2 .