The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=8ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $e^4$ is equal to

The values of $\lambda$,for which the point $(\lambda, \lambda-2)$ lies inside the ellipse $4x^2+9y^2=36$ and outside the parabola $y^2=x$,satisfy:

The equation of the line passing through the points of intersection of the parabola $x^2 = 8y$ and the ellipse $\frac{x^2}{3} + y^2 = 1$ is

Let $e_1$ and $e_2$ be the eccentricities of the ellipse $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$ and the hyperbola $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$,respectively. If $b < 5$ and $e_1 e_2 = 1$,then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$,with $y_1 < 0$ and $y_2 < 0$,be the endpoints of the latus rectum of the ellipse $x^2 + 4y^2 = 4$. The equations of the parabolas with latus rectum $PQ$ are:
$(A) x^2 + 2\sqrt{3}y = 3 + \sqrt{3}$
$(B) x^2 - 2\sqrt{3}y = 3 + \sqrt{3}$
$(C) x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$
$(D) x^2 - 2\sqrt{3}y = 3 - \sqrt{3}$

Find the equation of the ellipse whose focus is $(-1, 1)$,eccentricity is $1/2$,and directrix is $x - y + 3 = 0$.