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 \(\sqrt{5}\).

Leave a Comment

Your email address will not be published.