Question

Find the differential equation of the ellipse whose major axis is twice its minor axis.

Answer 1

Let 2a and 2b be lengths of major axis and minor axis of the ellipse.

Then 2a = 2(2b)

∴ a = 2b

∴ equation of the ellipse is

${\displaystyle \frac{{\text{x}}^2}{{\text{a}}^2}+\frac{{\text{y}}^2}{{\text{b}}^2}=1}$

i.e. ${\displaystyle \frac{{\text{x}}^2}{{\left(2\text{b}\right)}^2}+\frac{{\text{y}}^2}{{\text{b}}^2}=1}$

∴ ${\displaystyle \frac{{\text{x}}^2}{4{\text{b}}^2}+\frac{{\text{y}}^2}{{\text{b}}^2}=1}$

∴ x² + 4y² = 4b² 

Differentiating w.r.t. x, we get

${\displaystyle \text{2x}+4\times \text{2y}\frac{\text{dy}}{\text{dx}}=0}$

∴ ${\displaystyle \text{x}+\text{4y}\frac{\text{dy}}{\text{dx}}=0}$

This is the required D.E.