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10-2. Parabola, Ellipse, Hyperbola
easy
शांकव $16{x^2} + 7{y^2} = 112$ की उत्केन्द्रता है
A
$3\over \sqrt 7 $
B
$7\over{16}$
C
$3\over4$
D
$4\over3$
Solution
(c) $\frac{{{x^2}}}{{\frac{{112}}{{16}}}} + \frac{{{y^2}}}{{\frac{{112}}{7}}} = 1$
इसलिए $e = \sqrt {1 – \frac{{112}}{{16}}.\frac{7}{{112}}} = \frac{3}{4}$.
Standard 11
Mathematics