The eccentricity of the ellipse x2-16-y2-25-1

The correct option is D 3/4 x^216+ y^225=1 a=4, b=5 ecentricity, e=√(b^2 )a^2a =√( 1 b ^ 2 / a ^ 2 #160; ) =√( 1 25 / 16 #160; ) ...

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Standard XII

Mathematics

Definition of Ellipse

The eccentric...

Question

The eccentricity of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$

\frac{4}{5}

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\frac{3}{5}

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\frac{4}{3}

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\frac{3}{4}

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Solution

The correct option is D $\frac{3}{4}$
$\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$

$a = 4, b = 5$

ecentricity, $e = \frac{\sqrt{b^{2} -} a^{2}}{a}$

$= \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{25}{16}} = \frac{\sqrt{9}}{\sqrt{16}}$

$= \frac{3}{4}$

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The eccentricity of the ellipse x216+y225=1

The correct option is D 34x216+y225=1a=4, b=5ecentricity, e=√b2−a2a=√1−b2a2=√1−2516=√9√16=34

The eccentricity of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$

The correct option is D $\frac{3}{4}$
$\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$

$a = 4, b = 5$

ecentricity, $e = \frac{\sqrt{b^{2} -} a^{2}}{a}$

$= \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{25}{16}} = \frac{\sqrt{9}}{\sqrt{16}}$

$= \frac{3}{4}$