Find the coordinates of the foci, and the vertices, the eccentric… - Vidyadip

Question

Find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. $16x^2 - 9y^2 = 576$

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Answer

The given equation of hyperbola is $16x^2 - 9y^2 = 576$
i.e. $\frac{{16{x^2}}}{{576}} - \frac{{9{y^2}}}{{576}} = 1 \Rightarrow \frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{64}} = 1$ which is of the form $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$
The foci and vertices of the hyperbola lie on x-axis.
$\therefore a^2 = 36 \Rightarrow a = 6 and b^2 = 64 \Rightarrow b = 8$
Now $c^2 = a^2 + b^2 = 36 + 64 = 100 \Rightarrow$ c = 10
$\therefore$ Coordinates of foci are $( \pm c,0)$ i.e. $( \pm 10,0)$
Coordinates of vertices are $( \pm a,0)$ i.e. $( \pm 6,0)$
Eccentricity (e) $ = \frac{c}{a} = \frac{{10}}{6} = \frac{5}{3}$
Length of latus rectum $= \frac{{2{b^2}}}{a} = \frac{{2 \times 64}}{6} = \frac{{64}}{3}$

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