The distance between the directrices of the h | vedclass

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Question

(c) Equation of hyperbola is

$x = 8\,\sec 	heta ,\,y = 8	an 	heta $

==> $\frac{x}{8} = \sec 	heta ,\,\frac{y}{8} = 	an 	heta $

$\bec

Vedclass · Accepted Answer

$8\sqrt 2 $

Vedclass · Answer

(c) Equation of hyperbola is

$x = 8\,\sec 	heta ,\,y = 8	an 	heta $

==> $\frac{x}{8} = \sec 	heta ,\,\frac{y}{8} = 	an 	heta $

$\because \,\,{\sec ^2}	heta  - {	an ^2}	heta  = 1$ 

==> $\frac{{{x^2}}}{{{8^2}}} - \frac{{{y^2}}}{{{8^2}}} = 1$.

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 	imes 8}}{{\sqrt 2 }} = 8\sqrt 2 .$

The distance between the directrices of the hyperbola $x = 8\sec \theta ,\;\;y = 8\tan \theta $ is

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