Hyperbola practice · Easy

Let $\mathrm{H}$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latusrectum is _____.

If angle subtended by any chord of a rectangular hyperbola at the centre is $\alpha$ and angle between the tangents at ends of chord is $\beta$, then

The distance between the directrices of the hyperbola $x = 8\sec \mathrm{\theta },y = 8\mathrm{tan \theta }$ is

The equation of the hyperbola whose vertices are at $\left( {5,0} \right)$ and $\left( { - 5,\;0} \right)$ and one of the directrices is $x = \frac{{25}}{7},$, is

The equation of the hyperbola whose foci are $(8,3),(0,3)$ and eccentricity $=\frac{4}{3}$, is

The equation $13\left[ {{{\left( {x - 1} \right)}^2} + {{\left( {y - 2} \right)}^2}} \right] = 3{\left( {2x + 3y - 2} \right)^2}$ represents

If $\frac{x^2}{a+7}+\frac{y^2}{5-a}=1$ represents a hyperbola, then

The eccentricity of the hyperbola $\frac{\sqrt{1999}}{3}\left(x^2-y^2\right)=1$ is

If the normal at $P$ to the rectangular hyperbola $x^2-y^2=4$ meets the axes of $x$ and $y$ in $G$ and $g$ respectively and $C$ is the centre of the hyperbola, then $2 P C=$

If the origin is shifted to the point $\left(3, \frac{7}{2}\right)$ and the axes are rotated through an angle $\theta$ in a clockwise direction, so that the equation of the given hyperbola changes to the standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $\theta$ is

If $\alpha+\beta=3 \pi$, then the chord joining the points $\alpha$ and $\beta$ for the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through

The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ does not have mutually $\perp r$ tangents, then possible values of eccentricities are
a) $\frac{3}{2}\qquad$  b) $\frac{8}{7}\qquad$ c) $\frac{5}{3}\qquad$ d) $\frac{7}{5}$

The eccentricity of the hyperbola $4 x^2-9 y^2=16$ is

Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latusrectum. If $\mathrm{C}$ is the focus of the hyperbola nearest to the point $A$, then the area of the triangle $A B C$, is

The equation of the normal at the point $(a\sec \mathrm{\theta },\;b\tan \mathrm{\theta })\;$ of the curve $\;{b^2}{x^2} - {a^2}{y^2} = {a^2}{b^2}\;$ is

The eccentricity of the curve $x^2-y^2=1$ is

The equation $x^2+4 x y+y^2+2 x+4 y+2=0$ represents

The eccentricity of the conic defined by $\left|\sqrt{(x-1)^2+(y-2)^2}-\sqrt{(x-5)^2+(y-5)^2}\right|=3$ is $\mathrm{K}$, then $3 \mathrm{~K}=$

If the slope of a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \sqrt{2}$, then the eccentricity $e$ of the hyperbola lies in the interval.

From a point on the line $y=x+c, c$ (parameter), tangents are drawn to the hyperbola $\frac{x^2}{2}-\frac{y^2}{1}=1$ such that chords of contact pass through a fixed point $\left(x_1, y_1\right)$. Then, $\frac{x_1}{y_1}$ is equal to

The eccentricity of the hyperbola $\frac{x^2}{16}-\frac{y^2}{25}=1$ is

The latusractum of the hyperbola $9{x^2} - 16{y^2} + 72x - 32 - 16 = 0$ is

$P$ is any point on the hyperbola $x^2-y^2=a^2$. If $F_1$ and $F_2$ are the foci of the hyperbola and $P F_1 \cdot P F_2=\lambda(O P)^2$, where $O$ is the origin, then $\lambda$ is equal to

If a hyperbola passes through the point $P(10,\,16)$ and it has vertices at $( \pm 6,\,0),$ then the equation of the normal to it at $P$ is

The equation of the tangent to the hyperbola $16 x^2-9 y^2=144$ at $(5,8 / 3)$, is

If the normal at the point '$t_1$' to the rectangular hyperbola $x y=c^2$ meets it again at the point '$t_2$', then

An ellipse intersects the hyperbola $2 x^2-2 y^2=1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

If $\frac{x^2}{\lambda+3}+\frac{y^2}{2-\lambda}=1$ represents a hyperbola, then

The locus of the point of intersection of the straight lines, $\mathrm{tx}-2 \mathrm{y}-3 \mathrm{t}=0$ $x-2 t y+3=0(t \in R)$, is :

The eccentricity of the hyperbola $5{x^2} - 4{y^2} + 20x + 8y = 4$ is