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# Find the equation of normal for the parabola $y^2=4ax$ in parametric form

$\begin{array}{1 1}(a)\;y+x=2at+at^3\\(b)\;y^2+x^2=2at+at^3\\(c)\;y+x=at+at^2\\(d)\;y+2x=at+at^3\end{array}$

Given equation of the parabola is $y^2=4ax$
Differentiating w.r.to $x$, we get $2y\large\frac{dy}{dx}$$=4a \Rightarrow\:\large\frac{dy}{dx}=\frac{2a}{y} Slope of tangent to this curve at (x_1,y_1) is \large\frac{dy}{dx}=\frac{2a}{y_1} Slope of normal=\large\frac{-1}{dy/dx}$$=-\frac{y_1}{2a}$
$\therefore$ For the parabola $y^2=4ax$ equation of normal at $(x_1,y_1)$ is
$(y-y_1)=-\large\frac{-y_1}{2a}$$(x-x_1) Parametric coordinates of the parabola is given by (x_1,y_1)=(at^2,2at) \therefore Eqn. of normal becomes (y-2at)=-\large\frac{2at}{2a}$$(x-at^2)$
$\Rightarrow\:y-2at=-x+at^3$
$\Rightarrow\:y+x=2at+at^3$