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

If the curves $y^2 = 6x$ and $9x^2 + by^2 = 16$ intersect each other at right angles,then the value of $b$ is

If for $\theta \in \left[-\frac{\pi}{3}, 0\right]$,the points $(x, y) = \left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $xy+\alpha x+\beta y+\gamma=0$,then $\alpha^2+\beta^2+\gamma^2$ is equal to:

The equations of the common tangents to the ellipse $x^2 + 4y^2 = 8$ and the parabola $y^2 = 4x$ are

For the hyperbola $H : x^{2} - y^{2} = 1$ and the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where $a > b > 0$,let $(1)$ the eccentricity of $E$ be the reciprocal of the eccentricity of $H$,and $(2)$ the line $y = \sqrt{\frac{5}{2}} x + K$ be a common tangent of $E$ and $H$. Then $4(a^{2} + b^{2})$ is equal to:

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ are the same as the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2 = \dots$