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10-2. Parabola, Ellipse, Hyperbola
easy
अतिपरवलय $4{x^2} - 9{y^2} = 16$ की उत्केन्द्रता है
A
$\frac{8}{3}$
B
$\frac{5}{4}$
C
$\frac{{\sqrt {13} }}{3}$
D
$\frac{4}{3}$
Solution
(c) दिए गए अतिपरवलय से, $\frac{{{x^2}}}{4} – \frac{{{y^2}}}{{(16/9)}} = 1$,
$a = 2,\,\,b = \frac{4}{3}$
हम जानते हैं ${b^2} = {a^2}({e^2} – 1)$
$\frac{{16}}{9} = 4({e^2} – 1)$
${e^2} = \frac{{13}}{9}$,
$\therefore e = \frac{{\sqrt {13} }}{3}$.
Standard 11
Mathematics
