Equation of Hyperbola
Hyperbola

120767 If \(p, q\) are the eccentricities of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and its conjugate hyperbola respectively, then the area of the square (in sq. units) formed by the points of intersections of the ellipse \(\frac{x^2}{p^2}+\frac{y^2}{q^2}=1\) and the pair of lines \(\mathbf{x}^2-\mathrm{y}^2=0\) is

1 4
2 \(\sqrt{2}\)
3 \(\frac{\sqrt{3}}{2}\)
4 16
Hyperbola

120768 If the latus rectum of a hyperbola subtends an angle of \(120^{\circ}\) at its centre, then its eccentricity is

1 \(\frac{\sqrt{3}+2}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}+\sqrt{5}}{2}\)
3 \(\frac{\sqrt{3}-\sqrt{2}}{3}\)
4 \(\frac{\sqrt{3}+\sqrt{7}}{2}\)
Hyperbola

120769 The lines \(x \cos \alpha+y \sin \alpha=P, \alpha \in R\) are chords of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{36}=1\) and they subtend a right angle at the centre of the hyperbola. The locus of the poles of these lines with respect to the given hyperbola is

1 \(x^2-16 y^2=108\)
2 \(16 x^2-y^2=108\)
3 \(16 \mathrm{x}^2+\mathrm{y}^2=108\)
4 \(x^2+16 y^2=108\)
Hyperbola

120770 The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}-\frac{4(y-1)^2}{45}=1\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], we get -, We know that,, Comparing the above equation of hyperbola with standard equation of hyperbola, \(a^2=9\), \(a=3\), \(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\), \(\mathrm{e}^2=1+\frac{9}{9}=1+1=2\), \(\mathrm{e}=\sqrt{2}\), Distance between directories of hyperbola, \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2 \times 3}{\sqrt{2}}\), \(=3 \sqrt{2}\), 1058. If the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{b^2}=1\) and the hyperbola \(\frac{x^2}{144}-\frac{y^2}{25}=\frac{1}{13}\) coincide, then the value of \(b^2\) is,

1 \(\frac{3}{\sqrt{5}}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{\sqrt{3}}\)
4 \(\frac{5}{2}\)
5 \(\frac{3}{2}\)
Hyperbola

120679 The foci of the hyperbola \(4 x^2-9 y^2-1=0\) are

1 \(( \pm \sqrt{13}, 0)\)
2 \(\left( \pm \frac{\sqrt{13}}{6}, 0\right)\)
3 \(\left(0, \pm \frac{\sqrt{13}}{6}\right)\)
4 None of these
Hyperbola

120767 If \(p, q\) are the eccentricities of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and its conjugate hyperbola respectively, then the area of the square (in sq. units) formed by the points of intersections of the ellipse \(\frac{x^2}{p^2}+\frac{y^2}{q^2}=1\) and the pair of lines \(\mathbf{x}^2-\mathrm{y}^2=0\) is

1 4
2 \(\sqrt{2}\)
3 \(\frac{\sqrt{3}}{2}\)
4 16
Hyperbola

120768 If the latus rectum of a hyperbola subtends an angle of \(120^{\circ}\) at its centre, then its eccentricity is

1 \(\frac{\sqrt{3}+2}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}+\sqrt{5}}{2}\)
3 \(\frac{\sqrt{3}-\sqrt{2}}{3}\)
4 \(\frac{\sqrt{3}+\sqrt{7}}{2}\)
Hyperbola

120769 The lines \(x \cos \alpha+y \sin \alpha=P, \alpha \in R\) are chords of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{36}=1\) and they subtend a right angle at the centre of the hyperbola. The locus of the poles of these lines with respect to the given hyperbola is

1 \(x^2-16 y^2=108\)
2 \(16 x^2-y^2=108\)
3 \(16 \mathrm{x}^2+\mathrm{y}^2=108\)
4 \(x^2+16 y^2=108\)
Hyperbola

120770 The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}-\frac{4(y-1)^2}{45}=1\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], we get -, We know that,, Comparing the above equation of hyperbola with standard equation of hyperbola, \(a^2=9\), \(a=3\), \(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\), \(\mathrm{e}^2=1+\frac{9}{9}=1+1=2\), \(\mathrm{e}=\sqrt{2}\), Distance between directories of hyperbola, \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2 \times 3}{\sqrt{2}}\), \(=3 \sqrt{2}\), 1058. If the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{b^2}=1\) and the hyperbola \(\frac{x^2}{144}-\frac{y^2}{25}=\frac{1}{13}\) coincide, then the value of \(b^2\) is,

1 \(\frac{3}{\sqrt{5}}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{\sqrt{3}}\)
4 \(\frac{5}{2}\)
5 \(\frac{3}{2}\)
Hyperbola

120679 The foci of the hyperbola \(4 x^2-9 y^2-1=0\) are

1 \(( \pm \sqrt{13}, 0)\)
2 \(\left( \pm \frac{\sqrt{13}}{6}, 0\right)\)
3 \(\left(0, \pm \frac{\sqrt{13}}{6}\right)\)
4 None of these
Hyperbola

120767 If \(p, q\) are the eccentricities of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and its conjugate hyperbola respectively, then the area of the square (in sq. units) formed by the points of intersections of the ellipse \(\frac{x^2}{p^2}+\frac{y^2}{q^2}=1\) and the pair of lines \(\mathbf{x}^2-\mathrm{y}^2=0\) is

1 4
2 \(\sqrt{2}\)
3 \(\frac{\sqrt{3}}{2}\)
4 16
Hyperbola

120768 If the latus rectum of a hyperbola subtends an angle of \(120^{\circ}\) at its centre, then its eccentricity is

1 \(\frac{\sqrt{3}+2}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}+\sqrt{5}}{2}\)
3 \(\frac{\sqrt{3}-\sqrt{2}}{3}\)
4 \(\frac{\sqrt{3}+\sqrt{7}}{2}\)
Hyperbola

120769 The lines \(x \cos \alpha+y \sin \alpha=P, \alpha \in R\) are chords of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{36}=1\) and they subtend a right angle at the centre of the hyperbola. The locus of the poles of these lines with respect to the given hyperbola is

1 \(x^2-16 y^2=108\)
2 \(16 x^2-y^2=108\)
3 \(16 \mathrm{x}^2+\mathrm{y}^2=108\)
4 \(x^2+16 y^2=108\)
Hyperbola

120770 The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}-\frac{4(y-1)^2}{45}=1\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], we get -, We know that,, Comparing the above equation of hyperbola with standard equation of hyperbola, \(a^2=9\), \(a=3\), \(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\), \(\mathrm{e}^2=1+\frac{9}{9}=1+1=2\), \(\mathrm{e}=\sqrt{2}\), Distance between directories of hyperbola, \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2 \times 3}{\sqrt{2}}\), \(=3 \sqrt{2}\), 1058. If the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{b^2}=1\) and the hyperbola \(\frac{x^2}{144}-\frac{y^2}{25}=\frac{1}{13}\) coincide, then the value of \(b^2\) is,

1 \(\frac{3}{\sqrt{5}}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{\sqrt{3}}\)
4 \(\frac{5}{2}\)
5 \(\frac{3}{2}\)
Hyperbola

120679 The foci of the hyperbola \(4 x^2-9 y^2-1=0\) are

1 \(( \pm \sqrt{13}, 0)\)
2 \(\left( \pm \frac{\sqrt{13}}{6}, 0\right)\)
3 \(\left(0, \pm \frac{\sqrt{13}}{6}\right)\)
4 None of these
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Hyperbola

120767 If \(p, q\) are the eccentricities of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and its conjugate hyperbola respectively, then the area of the square (in sq. units) formed by the points of intersections of the ellipse \(\frac{x^2}{p^2}+\frac{y^2}{q^2}=1\) and the pair of lines \(\mathbf{x}^2-\mathrm{y}^2=0\) is

1 4
2 \(\sqrt{2}\)
3 \(\frac{\sqrt{3}}{2}\)
4 16
Hyperbola

120768 If the latus rectum of a hyperbola subtends an angle of \(120^{\circ}\) at its centre, then its eccentricity is

1 \(\frac{\sqrt{3}+2}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}+\sqrt{5}}{2}\)
3 \(\frac{\sqrt{3}-\sqrt{2}}{3}\)
4 \(\frac{\sqrt{3}+\sqrt{7}}{2}\)
Hyperbola

120769 The lines \(x \cos \alpha+y \sin \alpha=P, \alpha \in R\) are chords of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{36}=1\) and they subtend a right angle at the centre of the hyperbola. The locus of the poles of these lines with respect to the given hyperbola is

1 \(x^2-16 y^2=108\)
2 \(16 x^2-y^2=108\)
3 \(16 \mathrm{x}^2+\mathrm{y}^2=108\)
4 \(x^2+16 y^2=108\)
Hyperbola

120770 The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}-\frac{4(y-1)^2}{45}=1\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], we get -, We know that,, Comparing the above equation of hyperbola with standard equation of hyperbola, \(a^2=9\), \(a=3\), \(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\), \(\mathrm{e}^2=1+\frac{9}{9}=1+1=2\), \(\mathrm{e}=\sqrt{2}\), Distance between directories of hyperbola, \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2 \times 3}{\sqrt{2}}\), \(=3 \sqrt{2}\), 1058. If the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{b^2}=1\) and the hyperbola \(\frac{x^2}{144}-\frac{y^2}{25}=\frac{1}{13}\) coincide, then the value of \(b^2\) is,

1 \(\frac{3}{\sqrt{5}}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{\sqrt{3}}\)
4 \(\frac{5}{2}\)
5 \(\frac{3}{2}\)
Hyperbola

120679 The foci of the hyperbola \(4 x^2-9 y^2-1=0\) are

1 \(( \pm \sqrt{13}, 0)\)
2 \(\left( \pm \frac{\sqrt{13}}{6}, 0\right)\)
3 \(\left(0, \pm \frac{\sqrt{13}}{6}\right)\)
4 None of these
Hyperbola

120767 If \(p, q\) are the eccentricities of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and its conjugate hyperbola respectively, then the area of the square (in sq. units) formed by the points of intersections of the ellipse \(\frac{x^2}{p^2}+\frac{y^2}{q^2}=1\) and the pair of lines \(\mathbf{x}^2-\mathrm{y}^2=0\) is

1 4
2 \(\sqrt{2}\)
3 \(\frac{\sqrt{3}}{2}\)
4 16
Hyperbola

120768 If the latus rectum of a hyperbola subtends an angle of \(120^{\circ}\) at its centre, then its eccentricity is

1 \(\frac{\sqrt{3}+2}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}+\sqrt{5}}{2}\)
3 \(\frac{\sqrt{3}-\sqrt{2}}{3}\)
4 \(\frac{\sqrt{3}+\sqrt{7}}{2}\)
Hyperbola

120769 The lines \(x \cos \alpha+y \sin \alpha=P, \alpha \in R\) are chords of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{36}=1\) and they subtend a right angle at the centre of the hyperbola. The locus of the poles of these lines with respect to the given hyperbola is

1 \(x^2-16 y^2=108\)
2 \(16 x^2-y^2=108\)
3 \(16 \mathrm{x}^2+\mathrm{y}^2=108\)
4 \(x^2+16 y^2=108\)
Hyperbola

120770 The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}-\frac{4(y-1)^2}{45}=1\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], we get -, We know that,, Comparing the above equation of hyperbola with standard equation of hyperbola, \(a^2=9\), \(a=3\), \(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\), \(\mathrm{e}^2=1+\frac{9}{9}=1+1=2\), \(\mathrm{e}=\sqrt{2}\), Distance between directories of hyperbola, \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2 \times 3}{\sqrt{2}}\), \(=3 \sqrt{2}\), 1058. If the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{b^2}=1\) and the hyperbola \(\frac{x^2}{144}-\frac{y^2}{25}=\frac{1}{13}\) coincide, then the value of \(b^2\) is,

1 \(\frac{3}{\sqrt{5}}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{\sqrt{3}}\)
4 \(\frac{5}{2}\)
5 \(\frac{3}{2}\)
Hyperbola

120679 The foci of the hyperbola \(4 x^2-9 y^2-1=0\) are

1 \(( \pm \sqrt{13}, 0)\)
2 \(\left( \pm \frac{\sqrt{13}}{6}, 0\right)\)
3 \(\left(0, \pm \frac{\sqrt{13}}{6}\right)\)
4 None of these