The order and degree of the differential equation of the family of ellipses having the same foci, are respectively :

Solution :
The equation of the family is $$\frac{{{x^2}}}{{{a^2} + \lambda }} + \frac{{{y^2}}}{{{b^2} + \lambda }} = 1$$     because they have the same foci $$\left( { \pm \sqrt {{a^2} + {b^2}} ,\,0} \right).$$
On differentiation,
$$\eqalign{ & \frac{{2x}}{{{d^2} + \lambda }} + \frac{{2y}}{{{b^2} + \lambda }}.\frac{{dy}}{{dx}} = 0 \cr & {\text{or }}\frac{x}{{{a^2} + \lambda }} + \frac{{yp}}{{{b^2} + \lambda }} = 0,\,\,\left( {p = \frac{{dy}}{{dx}}} \right) \cr & {\text{or }}x\left( {{b^2} + \lambda } \right) + yp\left( {{a^2} + \lambda } \right) = 0 \cr & \Rightarrow \lambda = \frac{{ - {b^2}x - {a^2}yp}}{{x + yp}} \cr} $$
$$\therefore $$ the differential equation is,
$$\eqalign{ & \frac{{{x^2}}}{{{a^2} + \frac{{ - {b^2}x - {a^2}yp}}{{x + yp}}}} + \frac{{{y^2}}}{{{b^2} + \frac{{ - {b^2}x - {a^2}yp}}{{x + yp}}}} = 1 \cr & {\text{or }}\frac{{x\left( {x + yp} \right)}}{{{a^2} - {b^2}}} + \frac{{y\left( {x + yp} \right)}}{{\left( {{b^2} - {a^2}} \right)p}} = 1 \cr} $$
So, order $$=1$$   and degree $$=2.$$