x^2-y^2=16 - Vidyadip

Question

$x^2-y^2=16$

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Answer

Given equation of the hyperbola is $x^2-y^2=16$$\therefore \frac{x^2}{16}-\frac{y^2}{16}=1$
Comparing this equation with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, we get
$ a ^2=16 \text { and } b ^2=16$
$\therefore a =4 \text { and } b =4$
$\therefore a = 4$ and $b = 4$
(i) Length of transverse axis = 2a = 2(4) = 8
Length of conjugate axis = 2b = 2(4) = 8
(ii) We know that
$e =\frac{\sqrt{a^2+b^2}}{a}$
$ =\frac{\sqrt{16+16}}{4} $
$ =\frac{\sqrt{32}}{4} $
$ =\frac{4 \sqrt{2}}{4} $
$ =\sqrt{2}$
Co-ordinates of foci are $S(ae, 0)$ and $S'(-ae, 0),$
i.e., $S(4\sqrt{2}, 0)$ and $S'(-4√2, 0)$
Equations of the directrices are $x= \pm \frac{a}{e}$
$\therefore x= \pm \frac{4}{\sqrt{2}}$
$\therefore x= \pm 2 \sqrt{2}$
(iv) Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(16)}{4}=8$
(v) Distance between foci = 2ae = 2(4)(√2) = 8√2
(vi) Distance between directrices $=\frac{2 a}{e}=\frac{2(4)}{\sqrt{2}}=4 \sqrt{2}$

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