If the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide, then the value of ${b^2}$ is

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $36 x^{2}+4 y^{2}=144$

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ and the parabola $x^2=4(y+b)$ are such that the two foci of the ellipse and the end points of the latusrectum of parabola are the vertices of a square. The eccentricity of the ellipse is

In an ellipse, the distance between its foci is $6$ and minor axis is $8.$ Then its eccentricity is :

Locus of the foot of the perpendicular drawn from the centre upon any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, is

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