Correct Answer - Option 1 :
\((0, ± \sqrt{13})\)
Concept:
Hyperbola:
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Equation
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\(\rm \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
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\(\rm\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\)
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Equation of Transverse axis
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y = 0
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x = 0
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Equation of Conjugate axis
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x = 0
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y = 0
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Length of Transverse axis
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2a
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2a
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Length of Conjugate 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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\(\rm \sqrt{1+\frac{b^{2}}{a^{2}}}\)
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\(\rm \sqrt{1+\frac{b^{2}}{a^{2}}}\)
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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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ex ± a
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ey ± a
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Calculation:
Given: \(\rm\frac{y^{2}}{4}-\frac{x^{2}}{9}=1\)
It can be written as,
\(\rm⇒ \frac{y^{2}}{4}-\frac{x^{2}}{9}=\frac{y^{2}}{2^{2}}-\frac{x^{2}}{3^{2}}=1\)
On comparing the above equation with the general equation of the hyperbola, \(\rm\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\)
⇒ a = 2 and b = 3
As we know that the eccentricity of the hyperbola is \(\rm \sqrt{1+\frac{b^{2}}{a^{2}}}\).
\(\rm ⇒ \sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{3^{2}}{2^{2}}}=\sqrt{\frac{13}{4}}\)
As we know that the coordinates of the foci of the hyperbola \(\rm\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\) is (0, ± ae).
\(\rm ⇒(0, ± ae)=(0, ± 2\sqrt{\frac{13}{4}})=(0, ± \sqrt{13})\)
Hence, the correct option is 1.
