The foci of an ellipse are $(\pm 2, 0)$ and its eccentricity is 1/2, find its equation.

Solution :

Let the equation of the ellipse be $x^2\over a^2$ + $y^2\over b^2$ = 1.

Then, coordinates of the foci are $(\pm ae, 0)$.

Therefore,  ae = 2 $\implies$  a = 4

We have $b^2$ = $a^2(1 – e^2)$ $\implies$ $b^2$ =12

Thus, the equation of the ellipse is $x^2\over 16$ + $y^2\over 12$ = 1

---

Similar Questions

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

If the foci of a hyperbola are foci of the ellipse $x^2\over 25$ + $y^2\over 9$ = 1. If the eccentricity of the hyperbola be 2, then its equation is :

Find the equation of the tangents to the ellipse $3x^2+4y^2$ = 12 which are perpendicular to the line y + 2x = 4.

For what value of k does the line y = x + k touches the ellipse $9x^2 + 16y^2$ = 144.

Find the equation of ellipse whose foci are (2, 3), (-2, 3) and whose semi major axis is of length $\sqrt{5}$.