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\( =\frac{ 2b^2}{a} = \frac{(2 \times 72)}{4} = \frac{(2\times49)}{4} = \frac{49}{2}\)