On the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$ let $P$ be a point in the second quadrant such that the tangent at $\mathrm{P}$ to the ellipse is perpendicular to the line $x+2 y=0$. Let $S$ and $\mathrm{S}^{\prime}$ be the foci of the ellipse and $\mathrm{e}$ be its eccentricity. If $\mathrm{A}$ is the area of the triangle $SPS'$ then, the value of $\left(5-\mathrm{e}^{2}\right) . \mathrm{A}$ is :

If the points of intersection of two distinct conics $x^2+y^2=4 b$ and $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ lie on the curve $y^2=3 x^2$, then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is............................

Eccentricity of the ellipse $4{x^2} + {y^2} - 8x + 2y + 1 = 0$ is

Point $'O' $ is the centre of the ellipse with major axis $AB$ $ \&$ minor axis $CD$. Point $F$ is one focus of the ellipse. If $OF = 6 $  $ \&$  the diameter of the inscribed circle of triangle $OCF$  is $2, $ then the product $ (AB)\,(CD) $ is equal to

The locus of the mid point of the line segment joining the point $(4,3)$ and the points on the ellipse $x^{2}+2 y^{2}=4$ is an ellipse with eccentricity

If the line $y = mx + c$touches the ellipse $\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$, then $c = $