# 22) Find the equations of the tangent and normal to the parabola $y ^2 = 4 ax$ at the point $(at ^2, 2at).$

G Gautam harsolia

Equation of the given curve is
$y ^2 = 4 ax$

Slope of tangent = $2y\frac{dy}{dx} = 4a \Rightarrow \frac{dy}{dx} = \frac{4a}{2y}$
at point $(at ^2, 2at).$
$\frac{dy}{dx}= \frac{4a}{2(2at)} = \frac{4a}{4at} = \frac{1}{t}$
Now, the equation of tangent with point $(at ^2, 2at).$ and slope $\frac{1}{t}$ is
$y-y_1=m(x-x_1)\\ y-2at=\frac{1}{t}(x-at^2)\\ yt - 2at^2 = x - at^2\\ x-yt +at^2 = 0$

We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = -t$
Now, the equation of at point  $(at ^2, 2at).$  with slope -t
$y-y_1=m(x-x_1)\\ y-2at=(-t)(x-at^2)\\ y - 2at = -xt + at^3\\ xt+y -2at -at^3 = 0$

Hence, the equations of the tangent and normal to the parabola

$y ^2 = 4 ax$ at the point $(at ^2, 2at).$ are
$x-yt+at^2=0\ \ \ \ and \ \ \ \ xt+y -2at -at^3 = 0 \ \ respectively$

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