The eccentricity of an ellipse,with its centre at the origin,is $\frac{1}{2}$. If one of the directrices is $x = 4$,then the equation of the ellipse is

For the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,a chord $PQ$ subtends a right angle at the center. What is the locus of the point of intersection of the tangents at $P$ and $Q$?

Let $A, A^{\prime}$ be the end points of the major axis,$S, S^{\prime}$ be the foci,and $B, B^{\prime}$ be the end points of the minor axis of an ellipse $E$. If $\angle BAB^{\prime}=60^{\circ}$,then find $\angle SBS^{\prime}$.

The equation of an ellipse in its standard form,given the distance between its foci is $2$ units and the length of its latus rectum is $\frac{15}{2}$ units,is

Let $A_1$ be the area of the given ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let $A_2$ be the area of the region bounded by the curve which is the locus of the midpoint of the line segment joining the focus of the ellipse and a point $P$ on the given ellipse. Then $A_1 : A_2$ is equal to:

The chord joining two points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ subtends a right angle at the . . . point. (Given $\tan \theta_1 \tan \theta_2 = -\frac{a^2}{b^2}$)