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10-2. Parabola, Ellipse, Hyperbola
medium
The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is
A
$\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{144}} = 1$
B
$\frac{{{{(x - 5)}^2}}}{{25}} - \frac{{{y^2}}}{{144}} = 1$
C
$\frac{{{x^2}}}{{25}} - \frac{{{{(y - 5)}^2}}}{{144}} = 1$
D
$\frac{{{{(x - 5)}^2}}}{{25}} - \frac{{{{(y - 5)}^2}}}{{144}} = 1$
Solution
(b) $2a = 10$, $a = 5$
$ae – a = 8$ or $e = 1 + \frac{8}{5} = \frac{{13}}{5}$
$b = 5\sqrt {\frac{{{{13}^2}}}{{{5^2}}} – 1} = 5 \times \frac{{12}}{5} = 12$
and centre of hyperbola$ \equiv (5,\,0)$
$\frac{{{{(x – 5)}^2}}}{{{5^2}}} – \frac{{{{(y – 0)}^2}}}{{{{12}^2}}} = 1$.
Standard 11
Mathematics
