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

Let the circle $C$ touch the line $x - y + 1 = 0$,have the centre on the positive $x$-axis,and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3x + 2y = 1$. Let $H$ be the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$,whose one of the foci is the centre of $C$ and the length of the transverse axis is the diameter of $C$. Then $2\alpha^2 + 3\beta^2$ is equal to . . . . . .

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left( \frac{1}{2}, 1 \right)$. Its one directrix is the common tangent nearer to the point $P$, to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$. The equation of the ellipse in the standard form is:

The eccentricity of the conic $\frac{5}{r}=2+3 \cos \theta+4 \sin \theta$ is

$A$ line perpendicular to the $X$-axis cuts the circle $x^2+y^2=9$ at $A$ and the ellipse $4x^2+9y^2=36$ at $B$ such that $A$ and $B$ lie in the same quadrant. If $\theta$ is the greatest acute angle between the tangents drawn to the curves at $A$ and $B$,then $\tan \theta=$

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 $x$-axis and passing through $(0,2)$ is: