The distance between the foci of the ellipse | vedclass

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Question

(b) $\frac{{{x^2}}}{{(48/3)}} + \frac{{{y^2}}}{{(48/4)}} = 1$

${a^2} = 16,\;{b^2} = 12$

$e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \frac{1}{2}$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

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

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

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