Question

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $49y^ 2 - 16x ^2 = 784$

Answer 1

The given equation is $49y^ 2 - 16x ^2 = 784$
It can be written as $49y^ 2 - 16x^ 2 = 784$

or, $\frac{y^2}{16} –\frac{ x^2}{49} = 1$

or, $\frac{y^2}{42} – \frac{x^2}{72} = 1 .....(1)$

On comparing equation (1) with the standard equation of hyperbola i.e., $\frac{y^2}{a^2} –\frac{ x^2}{b^2} = 1$, we obtain a = 4 and b = 7

We know that $a^ 2 + b^ 2 = c^ 2 .$

$∴ c^2 = \frac{36}{5} + 4$

$c^2 = 16 + 49$

$c^2 = 65$

$⇒ c = \sqrt{65}$

Therefore,
The coordinates of the foci are (0, $\sqrt{65}$) and (0, –$\sqrt{65}$).
The coordinates of the vertices are (0, 4) and (0, -4).

Eccentricity, $e =\frac{ c}{a} = \frac{\sqrt{65}}{4}$

Length of latus rectum$=\frac{ 2b^2}{a} = \frac{(2 \times 72)}{4} = \frac{(2\times49)}{4} = \frac{49}{2}$