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Equation of normal for the parabola $y^2=4ax$ at point $(x_1,y_1)$ is

$\begin{array}{1 1}(a)\;(y-y_1)=-\large\frac{y_1}{2a}\normalsize (x-x_1)\\(b)\;(y-2y_1)=-\large\frac{y_1}{a}\normalsize (x-2x_1)\\(c)\;(y-3y_1)=-\large\frac{y_1}{a}\normalsize (x-x_1)\\(d)\;(y-y_1)=-\large\frac{y_1}{2a^2}\normalsize (x-x_1)\end{array}$

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A)
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For the parabola $y^2=4ax$
Slope of tangent at $(x_1,y_1)$ is
$2yy'=4a$
$y'=\large\frac{4a}{2y}$
$y'=\large\frac{2a}{y}$
$y'=\large\frac{2a}{y_1}$(at point $(x_1,y_1)$
Hence slope of normal is $-\large\frac{-y_1}{2a}$
Equation of normal $(y-y_1)=-\large\frac{-y_1}{2a}$$(x-x_1)$
Hence (a) is the correct answer.
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