Question

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

Answer 1

$(x-α)^2+2y^2=α^2$

$⇒x^2+α^2-2αx+2y^2=α^2$

$⇒2y^2=2αx-x^2$ ...(1)

Differentiating with respect to $x$, we get:

$2y\frac {dy}{dx} =\frac {2α-2x}{2}$

$⇒\frac {dy}{dx}=\frac {α-x}{2y}$

$⇒\frac {dy}{dx} = \frac {2αx-2x^2}{4xy}$ ...(2)

From equation(1), we get:

$2αx=2y^2+x^2$

On substituting this value in equation (3), we get:

$\frac {dy}{dx}=\frac {2y^2+x^2-2x^2}{4xy}$

$⇒\frac {dy}{dx} =\frac {2y^2-x^2}{4xy}$

Hence, the differential equation of the family of curves is given as $\frac {dy}{dx} =\frac {2y^2-x^2}{4xy}$.