Equation of Hyperbola
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NEET DIGITAL LIBRARY
Hyperbola

120759 If \((1,2)\) is the focus, \(x+2 y=0\) is the directrix and \(\sqrt{2}\) is the eccentricity of a hyperbola then the equation of the hyperbola is

1 \(x^2-y^2=a^2\)
2 \(3 x^2-8 x y-3 y^2-10 x-20 y+25=0\)
3 \(x y=c^2\)
4 \(3 x^2-8 x y-3 y^2+10 x-20 y-25=0\)
AP EAMCET-07.07.2022
Hyperbola

120760 A hyperbola having its centre at the origin is passing through the point \((5,2)\) and has transverse axis of length 8 along the \(\mathrm{X}\)-axis. Then the eccentricity of its conjugate hyperbola is

1 \(\frac{\sqrt{13}}{3}\)
2 \(\sqrt{\frac{13}{3}}\)
3 \(\frac{\sqrt{13}}{2}\)
4 \(\sqrt{\frac{13}{2}}\)
TS EAMCET-19.07.2022
Hyperbola

120761 The value of \(b^2\) in order that the foci of the hyperbola \(\frac{\mathrm{x}^2}{144}-\frac{\mathrm{y}^2}{81}=\frac{1}{25}\) and the ellipse \(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) coincide is

1 1
2 5
3 7
4 9
TS EAMCET-09.09.2020
Hyperbola

120762 If \(e_1\) is the eccentricity of the ellipse \(\frac{x^2}{16}+\frac{y^2}{25}=1\) and \(e_2\) is the eccentricity of a hyperbola passing through the foci of the given ellipse and \(e_1 e_2=1\), then the equation of such a hyperbola among the following is

1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
TS EAMCET-03.05.2019
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Hyperbola

120759 If \((1,2)\) is the focus, \(x+2 y=0\) is the directrix and \(\sqrt{2}\) is the eccentricity of a hyperbola then the equation of the hyperbola is

1 \(x^2-y^2=a^2\)
2 \(3 x^2-8 x y-3 y^2-10 x-20 y+25=0\)
3 \(x y=c^2\)
4 \(3 x^2-8 x y-3 y^2+10 x-20 y-25=0\)
AP EAMCET-07.07.2022
Hyperbola

120760 A hyperbola having its centre at the origin is passing through the point \((5,2)\) and has transverse axis of length 8 along the \(\mathrm{X}\)-axis. Then the eccentricity of its conjugate hyperbola is

1 \(\frac{\sqrt{13}}{3}\)
2 \(\sqrt{\frac{13}{3}}\)
3 \(\frac{\sqrt{13}}{2}\)
4 \(\sqrt{\frac{13}{2}}\)
TS EAMCET-19.07.2022
Hyperbola

120761 The value of \(b^2\) in order that the foci of the hyperbola \(\frac{\mathrm{x}^2}{144}-\frac{\mathrm{y}^2}{81}=\frac{1}{25}\) and the ellipse \(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) coincide is

1 1
2 5
3 7
4 9
TS EAMCET-09.09.2020
Hyperbola

120762 If \(e_1\) is the eccentricity of the ellipse \(\frac{x^2}{16}+\frac{y^2}{25}=1\) and \(e_2\) is the eccentricity of a hyperbola passing through the foci of the given ellipse and \(e_1 e_2=1\), then the equation of such a hyperbola among the following is

1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
TS EAMCET-03.05.2019
For More Updates join with Us
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Hyperbola

120759 If \((1,2)\) is the focus, \(x+2 y=0\) is the directrix and \(\sqrt{2}\) is the eccentricity of a hyperbola then the equation of the hyperbola is

1 \(x^2-y^2=a^2\)
2 \(3 x^2-8 x y-3 y^2-10 x-20 y+25=0\)
3 \(x y=c^2\)
4 \(3 x^2-8 x y-3 y^2+10 x-20 y-25=0\)
AP EAMCET-07.07.2022
Hyperbola

120760 A hyperbola having its centre at the origin is passing through the point \((5,2)\) and has transverse axis of length 8 along the \(\mathrm{X}\)-axis. Then the eccentricity of its conjugate hyperbola is

1 \(\frac{\sqrt{13}}{3}\)
2 \(\sqrt{\frac{13}{3}}\)
3 \(\frac{\sqrt{13}}{2}\)
4 \(\sqrt{\frac{13}{2}}\)
TS EAMCET-19.07.2022
Hyperbola

120761 The value of \(b^2\) in order that the foci of the hyperbola \(\frac{\mathrm{x}^2}{144}-\frac{\mathrm{y}^2}{81}=\frac{1}{25}\) and the ellipse \(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) coincide is

1 1
2 5
3 7
4 9
TS EAMCET-09.09.2020
Hyperbola

120762 If \(e_1\) is the eccentricity of the ellipse \(\frac{x^2}{16}+\frac{y^2}{25}=1\) and \(e_2\) is the eccentricity of a hyperbola passing through the foci of the given ellipse and \(e_1 e_2=1\), then the equation of such a hyperbola among the following is

1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
TS EAMCET-03.05.2019
NEET DIGITAL LIBRARY
Hyperbola

120759 If \((1,2)\) is the focus, \(x+2 y=0\) is the directrix and \(\sqrt{2}\) is the eccentricity of a hyperbola then the equation of the hyperbola is

1 \(x^2-y^2=a^2\)
2 \(3 x^2-8 x y-3 y^2-10 x-20 y+25=0\)
3 \(x y=c^2\)
4 \(3 x^2-8 x y-3 y^2+10 x-20 y-25=0\)
AP EAMCET-07.07.2022
Hyperbola

120760 A hyperbola having its centre at the origin is passing through the point \((5,2)\) and has transverse axis of length 8 along the \(\mathrm{X}\)-axis. Then the eccentricity of its conjugate hyperbola is

1 \(\frac{\sqrt{13}}{3}\)
2 \(\sqrt{\frac{13}{3}}\)
3 \(\frac{\sqrt{13}}{2}\)
4 \(\sqrt{\frac{13}{2}}\)
TS EAMCET-19.07.2022
Hyperbola

120761 The value of \(b^2\) in order that the foci of the hyperbola \(\frac{\mathrm{x}^2}{144}-\frac{\mathrm{y}^2}{81}=\frac{1}{25}\) and the ellipse \(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) coincide is

1 1
2 5
3 7
4 9
TS EAMCET-09.09.2020
Hyperbola

120762 If \(e_1\) is the eccentricity of the ellipse \(\frac{x^2}{16}+\frac{y^2}{25}=1\) and \(e_2\) is the eccentricity of a hyperbola passing through the foci of the given ellipse and \(e_1 e_2=1\), then the equation of such a hyperbola among the following is

1 \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
2 \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
3 \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
TS EAMCET-03.05.2019
Join With Us for More Updates