$\begin{array}{1 1} (A)\;y' -2y = 0 \\(B)\;xy' -y = 0 \\ (C)xy' +2y = 0 \\ (D)xy' -2y = 0 \end{array} $

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+1 vote

+1 vote

- Equation of a parabola having vertex at origin at axis along the positive axis is $x^2 = 4ay $

Step 1:

Given: $ x^2 = 4ay$

Differentiating on both sides we get,

$2x = 4ay'$

dividing by 2 on both sides,

$x = 2ay'$

$a =\large\frac{ x}{2y'}$

Step 2:

Substituting in equ(1) we get

$x^2 = 4. [\large\frac{x}{2y'}] $$.y$

On simplifying and rearranging we get

$xy' -2y = 0$

This is the required equation.

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