Question

Find the equation of the ellipse with eccentricity $\frac{3}{4},$ foci on the y-axis, centre at the origin and passing through the point (6, 4).

Answer 1

Since the centre of the ellipse lies at the origin and its foci lie on the y-axis, it is a vertical ellipse.

Let its equation be, $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1....\left(i\right)$

Let $c^2=a^2-b^2.$

Now, $e=\frac{3}{4}\iff \frac{c}{a}=\frac{3}{4}\iff c=\frac{3}{4}a.$

$\therefore b^2=(a^2-c^2)=(a^2-\frac{9}{16}{a}^2)=\frac{7a^2}{16}.$

So, the equation of the ellipse is,

$\frac{\frac{x^2}{7a^2}}{16}+\frac{y^2}{a^2}=1\iff 16x^2+7y^2=7a^2....\left(ii\right)$

Since it passes through (6, 4), putting x = 6 and y = 4 in (ii), we get,

$7a^2=688.$

Hence, the required equation is $16x^2+7y^2=688.$