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10-2. Parabola, Ellipse, Hyperbola
hard
જો બિંદુ $(K, 2)$ માંથી પસાર થતાં અતિવલય $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{{b^2}}} = 1$ ની ઉત્કેન્દ્રતા $\frac{{\sqrt {13} }}{3},$ હોય તો $K^2$ =
A
$18$
B
$8$
C
$1$
D
$2$
(AIEEE-2012)
Solution
given hyperbola is
$\frac{{{x^2}}}{9} – \frac{{{y^2}}}{{{b^2}}} = 1$
Since this passes through $(K,2)$, therfore
$\frac{{{K^2}}}{9} – \frac{4}{{{b^2}}} = 1\,\,\,\,\,\,….\left( 1 \right)$
Also, given $e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} = \frac{{\sqrt {13} }}{3}$
$ \Rightarrow \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} = \frac{{\sqrt {13} }}{3} \Rightarrow 9 + {b^2} = 13$
$ \Rightarrow b = \pm 2$
Now, from $e{q^n}$ $(1)$, we have
$\frac{{{K^2}}}{9} – \frac{4}{4} = 1$ ($\because$ $b = \pm 2$)
$ \Rightarrow {K^2} = 18$
Standard 11
Mathematics
