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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Differential Equations

Different equation of all parabola with x axis as the axis and $(a,0)$ as the focus

$(a)\;y \frac{dy}{dx} =2a \\ (b)\;\bigg(\frac{dy}{dx}\bigg)^3+\bigg(\frac{dy}{dx}\bigg)^2 (2x-2 \theta)+y=0 \\ (c)\;\bigg(\frac{dy}{dx}\bigg)= \frac{y}{2} \\ (d)\; \frac{dy}{dx}+(x-a)y=2 $

1 Answer

Let pirectrix latus rectum$=4b$
equation of parabola,
$(y-0)^2 =4b (x-a+b)$
$2y \large\frac{dy}{dx} $$=4b$
$y \large\frac{dy}{dx} $$=2b$
$y= 2 \bigg(\large\frac{dy}{dx} \bigg) \bigg( x- a +\large\frac{dy}{dx}\bigg)$
$y= 2\large\frac{ xdy}{dx} $$-2a \large\frac{dy}{dx} $$+2 \bigg(\large\frac{dy}{dx}\bigg)^2$
$\bigg(\large\frac{dy}{dx}\bigg)= \frac{y}{2}$
Hence c is the correct answer.
answered Feb 3, 2014 by meena.p
 

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