Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$,be $\frac{1}{4}$. If this ellipse passes through the point $\left(-4 \sqrt{\frac{2}{5}}, 3\right)$,then $a^{2}+b^{2}$ is equal to

The product of the lengths of the perpendiculars drawn from the two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ to the tangent at any point on the ellipse is

$S=(-1, 1)$ is the focus,$2x-3y+1=0$ is the directrix corresponding to $S$,and $\frac{1}{2}$ is the eccentricity of an ellipse. If $(a, b)$ is the centre of the ellipse,then $3a+2b=$

If the length of the latus rectum of the ellipse $x^{2} + 4y^{2} + 2x + 8y - \lambda = 0$ is $4$,and $l$ is the length of its major axis,then $\lambda + l$ is equal to $......$

$A$ rod $AB$ of length $15 \ cm$ rests between two coordinate axes such that the end point $A$ lies on the $x-$axis and the end point $B$ lies on the $y-$axis. $A$ point $P(x, y)$ is taken on the rod such that $AP = 6 \ cm$. Show that the locus of $P$ is an ellipse.

The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x$-axis at $Q$. If $M$ is the midpoint of the line segment $PQ$,then the locus of $M$ intersects the latus rectums of the given ellipse at the points