Find the eccentricity and foci of the ellipse $4 x^{2} + 9 y^{2} = 144$ .

Open in App

Solution

Given equation is $4 x^{2} + 9 y^{2} = 144$
On dividing throughout by $144$ , we get
$\frac{4}{144} x^{2} + \frac{9}{144} y^{2} = 1$
$\Rightarrow \frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$
On comparing the above equation with $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , we get
$a = 6, b = 4$
Therefore, $c = \sqrt{a^{2} - b^{2}} = \sqrt{6^{2} - 4^{2}} = \sqrt{36 - 16} = \sqrt{20} = 2 \sqrt{5}$
Thus the coordinates of the foci are given by $(- 2 \sqrt{5}, 0)$ and $(2 \sqrt{5}, 0)$ .
Therefore, eccentricity $e = \frac{c}{a} = \frac{2 \sqrt{5}}{6} = \frac{\sqrt{5}}{3}$ .

Suggest Corrections

Similar questions

Find the lengths of the major and minor axes; coordinates of the vertices and the foci; the eccentricity and length of the latus rectum of the ellipse.

$4 x^{2} + 9 y^{2} = 144.$

Find the lengths of the major and minor axes, coordinates of the vertices and the foci, the eccentricity and length of the latus rectum of the ellipse $4 x^{2} + 9 y^{2} = 144.$

Q. The eccentricity of ellipse

4 x^{2} + 9 y^{2} = 36

Find the length of latusrectum of the ellipse $4 x^{2} + 9 y^{2} = 144$

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$4 x^{2} + 9 y^{2} = 36$

Join BYJU'S Learning Program

Find the eccentricity and foci of the ellipse 4x2+9y2=144.

Find the eccentricity and foci of the ellipse $4 x^{2} + 9 y^{2} = 144$ .