# Find the equation of tangent at $x_1,y_1)$ for the parabola $y^2=4ax$?

$\begin{array}{1 1}(a)\;2yy_1=2a(x+x_1)\\(b)\;yy_1=a(x+x_1)\\(c)\;yy_1=2a(x+x_1)\\(d)\;3yy_1=a(x+x_1)\end{array}$

Slope of tangent at $(x_1,y_1)$
$2yy'=4a$
$y'=\large\frac{2a}{y_1}$
$(y-y_1)=\large\frac{2a}{y_1}$$(x-x_1)$
$yy_1-y_1^2=2ax-2ax_1$
$yy_1-(4ax_1)-2ax-2ax_1$
$(y_1^2=4ax_1)$
$yy_1=2ax+2ax_1$
$yy_1=2a(x+x_1)$
Hence (c) is the correct answer.