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.

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