The equation of the tangent parallel to $y - x + 5 = 0$ drawn to $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ is
$x - y - 1 = 0$
$x - y + 2 = 0$
$x + y - 1 = 0$
$x + y + 2 = 0$
The vertices of a hyperbola $H$ are $(\pm 6,0)$ and its eccentricity is $\frac{\sqrt{5}}{2}$. Let $N$ be the normal to $H$ at a point in the first quadrant and parallel to the line $\sqrt{2} x + y =2 \sqrt{2}$. If $d$ is the length of the line segment of $N$ between $H$ and the $y$-axis then $d ^2$ is equal to $............$.
The eccentricity of the hyperbola conjugate to ${x^2} - 3{y^2} = 2x + 8$ is
A tangent to the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{2} = 1$ meets $x-$ axis at $P$ and $y-$ axis at $Q$. Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on
The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$ , meet $x-$ axis and $y-$ axis at $A$ and $B$ respectively. Then $(OA)^2 - (OB)^2$ , where $O$ is the origin, equals
If $e$ and $e’$ are the eccentricities of the ellipse $5{x^2} + 9{y^2} = 45$ and the hyperbola $5{x^2} - 4{y^2} = 45$ respectively, then $ee' = $