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# Differential equation of all parabola with x axis as the axis and 4a as latus rectum .

$(a)\;y \frac{dy}{dx} =2a \\ (b)\;y \frac{dy}{dx} =4a \\ (c)\;x \frac{dy}{dx} =2a \\ (d)\;x \frac{dy}{dx} =4a$

The equation of parabola with any horizontal axis of symmetry $y=k$
$(y-k)2=4p (x-h)$
given $k=0$ and $p=a$
$y^2= 4a(x-h)$
$2y \large\frac{dy}{dx} $$=4a y \large\frac{dy}{dx}$$=2a$
Hence a is the correct answer.