NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120724
The eccentricity of the hyperbola \(4 x^2-9 y^2-36\) is
1 \(\frac{\sqrt{11}}{3}\)
2 \(\frac{\sqrt{15}}{3}\)
3 \(\frac{\sqrt{13}}{3}\)
4 \(\frac{\sqrt{14}}{3}\)
Explanation:
C Given,
\(4 x^2-9 y^2-36=0\)
\(\frac{x^2}{9}-\frac{y^2}{4}=1\)
Here, \(\mathrm{a}^2=9\) and \(\mathrm{b}^2=4\)
Eccentricity,
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}} \quad \mathrm{e}=\sqrt{1+\frac{4}{9}} \quad \mathrm{e}=\frac{\sqrt{13}}{3}\)
WB JEE-2011
Hyperbola
120725
The transverse axis of a hyperbola is along the \(x\)-axis and its length is \(2 a\). The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is
1 \(6 x^2-y^2=3 a^2\)
2 \(x^2-3 y^2=3 a^2\)
3 \(x^2-6 y^2=3 a^2\)
4 \(3 \mathrm{x}^2-\mathrm{y}^2=3 \mathrm{a}^2\)
Explanation:
D Let ' \(e\) ' be the eccentricity, and length of conjugate axis be \(2 \mathrm{~b}\).
\(\because\) Vertex \((a, 0)\) bisects centre \((0,0)\) and focus \((a e, 0)\)
\(\therefore \quad \mathrm{a}=\frac{\mathrm{ae}+0}{2}\)
\(\mathrm{e}=2\)
\(\therefore\) Equation of hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{x^2}{a^2}-\frac{y^2}{3 a^2}=1\)
\(3 x^2-y^2=3 a^2\)
WB JEE-2012
Hyperbola
120726
The equation of hyperbola whose coordinates of the foci are \(( \pm 8,0)\) and the length of latusrectum is 24 units, is
120724
The eccentricity of the hyperbola \(4 x^2-9 y^2-36\) is
1 \(\frac{\sqrt{11}}{3}\)
2 \(\frac{\sqrt{15}}{3}\)
3 \(\frac{\sqrt{13}}{3}\)
4 \(\frac{\sqrt{14}}{3}\)
Explanation:
C Given,
\(4 x^2-9 y^2-36=0\)
\(\frac{x^2}{9}-\frac{y^2}{4}=1\)
Here, \(\mathrm{a}^2=9\) and \(\mathrm{b}^2=4\)
Eccentricity,
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}} \quad \mathrm{e}=\sqrt{1+\frac{4}{9}} \quad \mathrm{e}=\frac{\sqrt{13}}{3}\)
WB JEE-2011
Hyperbola
120725
The transverse axis of a hyperbola is along the \(x\)-axis and its length is \(2 a\). The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is
1 \(6 x^2-y^2=3 a^2\)
2 \(x^2-3 y^2=3 a^2\)
3 \(x^2-6 y^2=3 a^2\)
4 \(3 \mathrm{x}^2-\mathrm{y}^2=3 \mathrm{a}^2\)
Explanation:
D Let ' \(e\) ' be the eccentricity, and length of conjugate axis be \(2 \mathrm{~b}\).
\(\because\) Vertex \((a, 0)\) bisects centre \((0,0)\) and focus \((a e, 0)\)
\(\therefore \quad \mathrm{a}=\frac{\mathrm{ae}+0}{2}\)
\(\mathrm{e}=2\)
\(\therefore\) Equation of hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{x^2}{a^2}-\frac{y^2}{3 a^2}=1\)
\(3 x^2-y^2=3 a^2\)
WB JEE-2012
Hyperbola
120726
The equation of hyperbola whose coordinates of the foci are \(( \pm 8,0)\) and the length of latusrectum is 24 units, is
120724
The eccentricity of the hyperbola \(4 x^2-9 y^2-36\) is
1 \(\frac{\sqrt{11}}{3}\)
2 \(\frac{\sqrt{15}}{3}\)
3 \(\frac{\sqrt{13}}{3}\)
4 \(\frac{\sqrt{14}}{3}\)
Explanation:
C Given,
\(4 x^2-9 y^2-36=0\)
\(\frac{x^2}{9}-\frac{y^2}{4}=1\)
Here, \(\mathrm{a}^2=9\) and \(\mathrm{b}^2=4\)
Eccentricity,
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}} \quad \mathrm{e}=\sqrt{1+\frac{4}{9}} \quad \mathrm{e}=\frac{\sqrt{13}}{3}\)
WB JEE-2011
Hyperbola
120725
The transverse axis of a hyperbola is along the \(x\)-axis and its length is \(2 a\). The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is
1 \(6 x^2-y^2=3 a^2\)
2 \(x^2-3 y^2=3 a^2\)
3 \(x^2-6 y^2=3 a^2\)
4 \(3 \mathrm{x}^2-\mathrm{y}^2=3 \mathrm{a}^2\)
Explanation:
D Let ' \(e\) ' be the eccentricity, and length of conjugate axis be \(2 \mathrm{~b}\).
\(\because\) Vertex \((a, 0)\) bisects centre \((0,0)\) and focus \((a e, 0)\)
\(\therefore \quad \mathrm{a}=\frac{\mathrm{ae}+0}{2}\)
\(\mathrm{e}=2\)
\(\therefore\) Equation of hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{x^2}{a^2}-\frac{y^2}{3 a^2}=1\)
\(3 x^2-y^2=3 a^2\)
WB JEE-2012
Hyperbola
120726
The equation of hyperbola whose coordinates of the foci are \(( \pm 8,0)\) and the length of latusrectum is 24 units, is
120724
The eccentricity of the hyperbola \(4 x^2-9 y^2-36\) is
1 \(\frac{\sqrt{11}}{3}\)
2 \(\frac{\sqrt{15}}{3}\)
3 \(\frac{\sqrt{13}}{3}\)
4 \(\frac{\sqrt{14}}{3}\)
Explanation:
C Given,
\(4 x^2-9 y^2-36=0\)
\(\frac{x^2}{9}-\frac{y^2}{4}=1\)
Here, \(\mathrm{a}^2=9\) and \(\mathrm{b}^2=4\)
Eccentricity,
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}} \quad \mathrm{e}=\sqrt{1+\frac{4}{9}} \quad \mathrm{e}=\frac{\sqrt{13}}{3}\)
WB JEE-2011
Hyperbola
120725
The transverse axis of a hyperbola is along the \(x\)-axis and its length is \(2 a\). The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is
1 \(6 x^2-y^2=3 a^2\)
2 \(x^2-3 y^2=3 a^2\)
3 \(x^2-6 y^2=3 a^2\)
4 \(3 \mathrm{x}^2-\mathrm{y}^2=3 \mathrm{a}^2\)
Explanation:
D Let ' \(e\) ' be the eccentricity, and length of conjugate axis be \(2 \mathrm{~b}\).
\(\because\) Vertex \((a, 0)\) bisects centre \((0,0)\) and focus \((a e, 0)\)
\(\therefore \quad \mathrm{a}=\frac{\mathrm{ae}+0}{2}\)
\(\mathrm{e}=2\)
\(\therefore\) Equation of hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{x^2}{a^2}-\frac{y^2}{3 a^2}=1\)
\(3 x^2-y^2=3 a^2\)
WB JEE-2012
Hyperbola
120726
The equation of hyperbola whose coordinates of the foci are \(( \pm 8,0)\) and the length of latusrectum is 24 units, is
