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