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10-2. Parabola, Ellipse, Hyperbola
easy
The distance between the directrices of the hyperbola $x = 8\sec \theta ,\;\;y = 8\tan \theta $ is
A
$16\sqrt 2 $
B
$\sqrt 2 $
C
$8\sqrt 2 $
D
$4\sqrt 2 $
Solution
(c) Equation of hyperbola is
$x = 8\,\sec \theta ,\,y = 8\tan \theta $
==> $\frac{x}{8} = \sec \theta ,\,\frac{y}{8} = \tan \theta $
$\because \,\,{\sec ^2}\theta – {\tan ^2}\theta = 1$
==> $\frac{{{x^2}}}{{{8^2}}} – \frac{{{y^2}}}{{{8^2}}} = 1$.
Here, $a = 8,\,\,b = 8$
Now, $e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 + \frac{{{8^2}}}{{{8^2}}}} = \sqrt {1 + 1} $
==> $e = \sqrt 2 $
Distance between directrices $ = \frac{{2a}}{e}$$ = \frac{{2 \times 8}}{{\sqrt 2 }} = 8\sqrt 2 .$
Standard 11
Mathematics