If the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ having $(1,1)$ as its midpoint is $x+\alpha y=\beta$,then

In an ellipse,the distance from one of the foci to its corresponding end of the major axis is $4-\sqrt{7}$ and the distance from the same focus to one end of the minor axis is $4$. Then the cosine of the angle subtended by the line segment joining its foci at one end of its minor axis is

The locus of the midpoints of the segments of the tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intercepted between the coordinate axes is:

Consider ellipses $E_{k}: kx^{2} + k^{2}y^{2} = 1$,for $k = 1, 2, \ldots, 20$. Let $C_{k}$ be the circle which touches the four chords joining the end points (one on the minor axis and another on the major axis) of the ellipse $E_{k}$. If $r_{k}$ is the radius of the circle $C_{k}$,then the value of $\sum_{k=1}^{20} \frac{1}{r_{k}^{2}}$ is $.......$.

The locus of a point such that the sum of its distances from the points $(0, 2)$ and $(0, -2)$ is $6$ is:

Let a focus of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $S(4, 0)$ and its eccentricity be $\frac{4}{5}$. If the point $P(3, \alpha)$ lies on $E$ and $O$ is the origin, then the area of $\triangle POS$ is equal to: (in $/ 5$)