Let $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4 \sqrt{3}$. Suppose the point $(\alpha, 6), \alpha>0$ lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2+\beta$ is equal to :

The locus of the midpoints of the chord of the circle, $x^{2}+y^{2}=25$ which is tangent to the hyperbola $, \frac{ x ^{2}}{9}-\frac{ y ^{2}}{16}=1$ is

Let $e_1$ be the eccentricity of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and $e_2$ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$, which passes through the foci of the hyperbola. If $e_1 e_2=1$, then the length of the chord of the ellipse parallel to the $\mathrm{x}$-axis and passing through $(0,2)$ is :

Let $H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$, where $a > b >0$, be $a$ hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4 \sqrt{3}$..

The correct option is:

A point on the curve $\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}} = 1$ is

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$