Question

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \(16x^ 2 - 9y^2 = 576\)

Answer 1

The given equation is \(16x ^2 - 9y ^2 = 576.\) 
It can be written as \(16x^ 2 - 9y^ 2 = 576\)
or \(\frac{x^2}{36} –\frac{ y^2}{64} = 1\)

or \(\frac{x^2}{6^2} – \frac{y^2}{8^2} = 1.......(1)\)

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 = 6 and b = 8.
We know that \(a ^2 + b ^2 = c^ 2 .\)

\(∴ c^2 = 36 + 64\)
\(c = \sqrt{100}\)
\(c = 10\)

Therefore, 
The coordinates of the foci are (±10, 0).
The coordinates of the vertices are (±6, 0)

Eccentricity, \(e =\frac{ c}{a} = \frac{10}{6} = \frac{5}{3}\)

Length of latus rectum \(=\frac{ 2b^2}{a} = \frac{(2 \times 82)}{6} = \frac{(2\times64)}{6} = \frac{64}{3}\)