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

If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25x^{2} + 4y^{2} = 1$ to the parabola $y^{2} = 4x$ are such that the slope of one tangent is four times the other,then the value of $(10\alpha + 5)^{2} + (16\beta^{2} + 50)^{2}$ equals

Let $P$ be a moving point on the circle $x^2 + y^2 - 6x - 8y + 21 = 0$. Then, the maximum distance of $P$ from the vertex of the parabola $x^2 + 6x + y + 13 = 0$ is equal to:

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle,where $O$ is the centre of the hyperbola,then the eccentricity $e$ of the hyperbola satisfies:

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 $b^2 = \dots$

The point$(s)$ on the parabola $y^2 = 4x$ which are closest to the circle $x^2 + y^2 - 24y + 128 = 0$ is/are: