Find the equation of axis of the given hyperbola $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ which is equally inclined to the axes

The eccentricity of the hyperbola conjugate to ${x^2} - 3{y^2} = 2x + 8$ is

Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse, $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b<5)$ and the hyperbola $\frac{ x ^{2}}{16}-\frac{ y ^{2}}{ b ^{2}}=1$ respectively satisfying $e _{1} e _{2}=1 .$ If $\alpha$ and $\beta$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha, \beta)$ is equal to

Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola $16y^2 - 9x^2 = 1 $ is

The equation of a hyperbola, whose foci are $(5, 0)$ and $(-5, 0)$ and the length of whose conjugate axis is $8$, is

If the circle $x^2 + y^2 = a^2$  intersects the hyperbola $xy = c^2 $ in four points $ P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3), S(x_4, y_4), $ then