# Q3.    Form the differential equation representing the family of curves given by $(x-a)^2 + 2y^2 = a^2$ , where a is an arbitrary constant.

Given equation is
$(x-a)^2 + 2y^2 = a^2$
we can rewrite it as
$2y^2 = a^2-(x-a)^2$                   -(i)
Differentiate both the sides w.r.t x
$\frac{d\left ( 2y^2 \right )}{dx}=\frac{d(a^2-(x-a)^2)}{dx}$
$4yy^{'}=4y\frac{dy}{dx}=-2(x-a)\\ \\$
$(x-a)= -2yy'\Rightarrow a = x+2yy'$                          -(ii)
Put value from equation (ii)  in (i)
$(-2yy')^2+2y^2= (x+2yy')^2\\ 4y^2(y')^2+2y^2= x^2+4y^2(y')^2+4xyy'\\ y' = \frac{2y^2-x^2}{4xy}$
Therefore, the required differential equation is $y' = \frac{2y^2-x^2}{4xy}$

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