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10-2. Parabola, Ellipse, Hyperbola
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दीर्घवृत्त $\frac{{{{(x - 1)}^2}}}{9} + \frac{{{{(y + 1)}^2}}}{{25}} = 1$ की उत्केन्द्रता है
A
$4\over5$
B
$3\over5$
C
$5\over4$
D
अधिकल्पित
Solution
(a) ${a^2} = {b^2}(1 – {e^2})$ ,
$9 = 25(1 – {e^2})$
$\frac{9}{{25}} = 1 – {e^2}$
${e^2} = \frac{{16}}{{25}}$
$e = \frac{4}{5}$.
Standard 11
Mathematics