# Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas : $49y^2 – 16x^2 = 784$

$\begin {array} {1 1} (A)\;Foci : (\pm \sqrt{65}, 0), Vertices : (\pm 4, 0), e = \sqrt{65} / 4, Length \: of \: the \: latus \: rectum : 49 / 2 \\ (B)\;Foci : (0, \pm \sqrt{65}), Vertices : (0, \pm 4), e = \sqrt{65} / 4, Length \: of \: the \: latus \: rectum : 49 / 2 \\ (C)\;Foci : ( \sqrt{65}, 0), Vertices : ( 4, 0), e = \sqrt{65} / 4, Length \: of \: the \: latus \: rectum : 49 / 2 \\ (D)\;Foci : (0, - \sqrt{65}), Vertices : (0, - 4), e = \sqrt{65} / 4, Length \: of \: the \: latus \: rectum : 49 / 2 \end {array}$

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 :
The given equation is $49y^2-16x^2=784$
Dividing throughout by 784 we get,
$\large\frac{y^2}{16}$$-\large\frac{x^2}{49}$$=1$
Comparing this with the standard equation of hyperbola
$\large\frac{y^2}{a^2}$$-\large\frac{x^2}{b^2}$$=1$ we get,
$a^2=16 \Rightarrow a = \pm 4$
$b^2=49 \Rightarrow b \pm 7$
$\therefore c^2 = a^2+b^2$
(i.e) $c^2 = 16+49 = 65$
$\therefore c = \sqrt{65}$
Step 2 :
Hence coordinates of foci are : $( 0, \pm \sqrt{65} )$
Coordinates of vertices are : $( 0, \pm 4 )$
eccentricity is $\large\frac{c}{a}$$= \large\frac{\sqrt{65}}{4} Length of the latus rectum is \large\frac{2b^2}{a} = \large\frac{2 \times 49}{4}$$ = \large\frac{49}{2}$