Correct Answer - Option 4 : (0, ± 3)
Concept:
Ellipse:
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Equation
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\(\rm \frac{x^{2}}{a^2}+\frac{y^{2}}{b^2}=1\) (a > b)
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\(\rm \frac{x^{2}}{b^2}+\frac{y^{2}}{a^2}=1\) (a > b)
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Equation of Major axis
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y = 0
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x = 0
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Equation of Minor axis
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x = 0
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y = 0
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Length of Major axis
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2a
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2a
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Length of Minor axis
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2b
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2b
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Vertices
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(± a, 0)
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(0, ± a)
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Focus
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(± ae, 0)
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(0, ± ae)
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Directrix
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x = ± a/e
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y = ± a/e
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Centre
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(0, 0)
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(0, 0)
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Eccentricity
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\(\sqrt{1-\frac {b^2}{a^{2}}}\)
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\(\sqrt{1-\frac {b^2}{a^{2}}}\)
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Length of Latus rectum
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\(\rm \frac{2b^{2}}{a}\)
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\(\rm \frac{2b^{2}}{a}\)
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Focal distance of the point (x, y)
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a ± ex
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a ± ey
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Calculation:
Given: \(\rm \frac{x^{2}}{16}+\frac{y^{2}}{25}=1\)
It can be written in standard form as, \(\rm \frac{x^{2}}{4^{2}}+\frac{y^{2}}{5^{2}}=1\)
On comparing with the general equation of the ellipse, \(\rm \frac{x^{2}}{b^2}+\frac{y^{2}}{a^2}=1\) where a > b
⇒ a = 5 and b = 4
As we know that, the eccentricity of the ellipse is \(\rm \frac{x^{2}}{b^2}+\frac{y^{2}}{a^2}=1\) where a > b is \(\sqrt{1-\frac {b^2}{a^{2}}}\)
\(\rm ⇒ e= \sqrt{1-\frac {b^2}{a^{2}}}= \sqrt{1-\frac {4^2}{5^{2}}}=\sqrt{\frac {25-16}{25}}=\frac{{3}}{5}\)
As we know that, the focus of the ellipse \(\rm \frac{x^{2}}{b^2}+\frac{y^{2}}{a^2}=1\) where a > b is (0, ± ae).
So, the focus of the required ellipse is (0, ± 3)
Hence, the correct option is 4.
