Distance between foci of ellipse 3{x^2} + 4{y^2} = 48 is

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10-2. Parabola, Ellipse, Hyperbola

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The distance between the foci of the ellipse $3{x^2} + 4{y^2} = 48$ is

A

$2$

B

$4$

C

$6$

D

$8$

Solution

(b) $\frac{{{x^2}}}{{(48/3)}} + \frac{{{y^2}}}{{(48/4)}} = 1$

${a^2} = 16,\;{b^2} = 12$

$e = \sqrt {1 – \frac{{{b^2}}}{{{a^2}}}} = \frac{1}{2}$

Distance is $2ae = 2 \cdot 4 \cdot \frac{1}{2} = 4$.

Standard 11

Mathematics

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