Find the equation of the ellipse whose axes are along the coordinate axes, vertices are \((0, \pm 10)\) and eccentricity e = 4/5.

Solution :

Let the equation of the required ellipse be

\(x^2\over a^2\) + \(y^2\over b^2\) = 1 ……….(i)

Since the vertices of the ellipse are on y-axis.

So, the coordinates of the vertices are \((0, \pm b)\).

\(\therefore\) b = 10

Now, \(a^2\) = \(b^2(1 – e^2)\) \(\implies\) \(a^2\) = 100(1 – 16/25) = 36

Substituting the values of \(a^2\) and \(b^2\) in (i), we obtain

\(x^2\over 36\) + \(y^2\over 100\) = 1 as the required equation of the ellipse.