Browse Questions

# Find the equations of the hyperbola satisfying the given conditions : Vertices $(0, \pm 3),$ foci $(0, \pm 5)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{16}-\large\frac{y^2}{9}=1 & \quad (B)\;\large\frac{x^2}{9}-\large\frac{y^2}{16}=1 \\ (C)\;\large\frac{x^2}{16}+\large\frac{y^2}{9}=1 & \quad (D)\;\large\frac{x^2}{9}+\large\frac{y^2}{16}=1 \end {array}$

Toolbox:
• Standard equation of a hyperbola where transverse axis is along x - axis is $\large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2}$$=1$
• Standard equation of a hyperbola whose transverse axis is along y - axis is $\large\frac{y^2}{a^2} $$- \large\frac{x^2}{b^2}$$=1$
• $c^2=a^2+b^2$, where $c$ is the foci
• Eccentricity $e = \large\frac{c}{a}$
• Length of the latus rectum $= \large\frac{2b^2}{a}$
Step 1 :
Coordinates of vertices and foci are $(0, \pm 3 )$ and foci $(0, \pm 5)$ respectively.
Clearly the vertices are on the y - axis.
Hence the equation of the hyperbola is of the form $\large\frac{y^2}{a^2}$$- \large\frac{x^2}{b^2}$$=1$
Since the vertices are $(0, \pm 3)$
$\therefore a = 3$
Since the foci are $(0, \pm 5) \Rightarrow c = 5$
$\therefore c^2 = a^2+b^2$
$\Rightarrow b^2 = 25-9$
$= 16$
Hence the equation of the hyperbola is
$\large\frac{y^2}{9}$$- \large\frac{x^2}{16}$$=1$