Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas : $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1.$
Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas : $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1.$
Comparing the equation $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ with the standard equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Here, $a=3,\,\, b=4$ and $c=\sqrt{a^{2}+b^{2}}=\sqrt{9+16}=5$
Therefore, the coordinates of the foci are $(±5,\,0)$ and that of vertices are $(\pm 3,\,0) .$ Also,
The eccentricity $e=\frac{c}{a}=\frac{5}{3} .$
The latus rectum $=\frac{2 b^{2}}{a}=\frac{32}{3}$
Similar Questions
The reciprocal of the eccentricity of rectangular hyperbola, is
The reciprocal of the eccentricity of rectangular hyperbola, 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 latus rectum. If $\mathrm{C}$ is the focus of the hyperbola nearest to the point $\mathrm{A}$, then the area of the triangle $\mathrm{ABC}$ 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 latus rectum. If $\mathrm{C}$ is the focus of the hyperbola nearest to the point $\mathrm{A}$, then the area of the triangle $\mathrm{ABC}$ is
- [IIT 2008]
Let $P (3\, sec\,\theta , 2\, tan\,\theta )$ and $Q\, (3\, sec\,\phi , 2\, tan\,\phi )$ where $\theta + \phi \, = \frac{\pi}{2}$ , be two distinct points on the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1$ . Then the ordinate of the point of intersection of the normals at $P$ and $Q$ is
Let $P (3\, sec\,\theta , 2\, tan\,\theta )$ and $Q\, (3\, sec\,\phi , 2\, tan\,\phi )$ where $\theta + \phi \, = \frac{\pi}{2}$ , be two distinct points on the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1$ . Then the ordinate of the point of intersection of the normals at $P$ and $Q$ is
- [JEE MAIN 2014]
A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^2+4 y^2=12$. Then its equation is
A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^2+4 y^2=12$. Then its equation is
- [IIT 2007]
The latus-rectum of the hyperbola $16{x^2} - 9{y^2} = $ $144$, is
The latus-rectum of the hyperbola $16{x^2} - 9{y^2} = $ $144$, is