If ${m_1}$ and ${m_2}$ are the slopes of the tangents to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ which pass through the point $(6, 2)$,then:

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$,be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$,then $\beta$ is equal to

Let the foci of the hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$. If the eccentricity of the hyperbola is $5$, then the length of its latus rectum is:

The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2 - 2y^2 - 2 = 0$ to its asymptotes is:

The equation of the asymptotes of the hyperbola $2 x^2+5 x y+2 y^2-11 x-7 y-4=0$ is

Let the ellipse $E: \frac{x^{2}}{144}+\frac{y^{2}}{169}=1$ and the hyperbola $H: \frac{x^{2}}{16}-\frac{y^{2}}{\lambda^{2}}=-1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$, then the value of $24(e+L)$ is: