# the line lx + my +n=0 is a normal to the parabola y^2 = 4ax if

the equation of the given parabola is
the equation of the normal to the given parabola is
i.e. $tx+y-\left(2at+a{t}^{3}\right)=0$ ........(1)
the equation (1) and  the given equation of the normal  represent the same line:
therefore
$\frac{t}{l}=\frac{1}{m}=\frac{-\left(2at+a{t}^{3}\right)}{n}$

hope this helps you

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