Tangent and Normal to Hyperbola
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Hyperbola

120801 A tangent to the curve \(9 b^2 x^2-4 a^2 y^2=36 a^2 b^2\) makes intercepts of unit length on each of the coordinate axes, then the point \((a, b)\) lies on

1 \(x^2-y^2=1\)
2 \(x^2+y^2=1\)
3 \(4 x^2-9 y^2=1\)
4 \(4 x^2+9 y^2=1\)
AP EAMCET-23.04.2018
Hyperbola

120802 The ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and the hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) have in common

1 centre only
2 centre, foci and directrices
3 centre, foci and vertices
4 centre and vertices only
Karnataka CET-2006
Hyperbola

120803 The equation of those tangents to \(4 x^2-9 y^2=\) 36 which are perpendicular to the straight line \(5 x+2 y-10=0\) are

1 \(5(\mathrm{y}-3)=2\left(\mathrm{x}-\frac{\sqrt{117}}{2}\right)\)
2 \(2 x-5 y+10-2 \sqrt{18}=0\)
3 \(2 \mathrm{x}-5 \mathrm{y}-10-2 \sqrt{18}=0\)
4 None of the above
Jamia Millia Islamia-2009
Hyperbola

120804 The eccentricity of the hyperbola \(x^2 / 49-y^2 / 576\) \(=1\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: And,

1 \(25 / 24\)
2 \(25 / 7\)
3 \(1 / 25\)
4 \(49 / 576\)
And
Hyperbola

120805 Tangents are drawn to the hyperbola \(4 x^2-y^2=\) 36 at the points \(P\) and \(Q\). If these tangents intersect at the point \(T(0,3)\), then the area (in sq units) of \(\triangle \mathrm{PTQ}\) is

1 \(45 \sqrt{5}\)
2 \(54 \sqrt{3}\)
3 \(60 \sqrt{3}\)
4 \(36 \sqrt{5}\)
JEE Main-2018
NEET DIGITAL LIBRARY
Hyperbola

120801 A tangent to the curve \(9 b^2 x^2-4 a^2 y^2=36 a^2 b^2\) makes intercepts of unit length on each of the coordinate axes, then the point \((a, b)\) lies on

1 \(x^2-y^2=1\)
2 \(x^2+y^2=1\)
3 \(4 x^2-9 y^2=1\)
4 \(4 x^2+9 y^2=1\)
AP EAMCET-23.04.2018
Hyperbola

120802 The ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and the hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) have in common

1 centre only
2 centre, foci and directrices
3 centre, foci and vertices
4 centre and vertices only
Karnataka CET-2006
Hyperbola

120803 The equation of those tangents to \(4 x^2-9 y^2=\) 36 which are perpendicular to the straight line \(5 x+2 y-10=0\) are

1 \(5(\mathrm{y}-3)=2\left(\mathrm{x}-\frac{\sqrt{117}}{2}\right)\)
2 \(2 x-5 y+10-2 \sqrt{18}=0\)
3 \(2 \mathrm{x}-5 \mathrm{y}-10-2 \sqrt{18}=0\)
4 None of the above
Jamia Millia Islamia-2009
Hyperbola

120804 The eccentricity of the hyperbola \(x^2 / 49-y^2 / 576\) \(=1\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: And,

1 \(25 / 24\)
2 \(25 / 7\)
3 \(1 / 25\)
4 \(49 / 576\)
And
Hyperbola

120805 Tangents are drawn to the hyperbola \(4 x^2-y^2=\) 36 at the points \(P\) and \(Q\). If these tangents intersect at the point \(T(0,3)\), then the area (in sq units) of \(\triangle \mathrm{PTQ}\) is

1 \(45 \sqrt{5}\)
2 \(54 \sqrt{3}\)
3 \(60 \sqrt{3}\)
4 \(36 \sqrt{5}\)
JEE Main-2018
Join With Us for More Updates
Hyperbola

120801 A tangent to the curve \(9 b^2 x^2-4 a^2 y^2=36 a^2 b^2\) makes intercepts of unit length on each of the coordinate axes, then the point \((a, b)\) lies on

1 \(x^2-y^2=1\)
2 \(x^2+y^2=1\)
3 \(4 x^2-9 y^2=1\)
4 \(4 x^2+9 y^2=1\)
AP EAMCET-23.04.2018
Hyperbola

120802 The ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and the hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) have in common

1 centre only
2 centre, foci and directrices
3 centre, foci and vertices
4 centre and vertices only
Karnataka CET-2006
Hyperbola

120803 The equation of those tangents to \(4 x^2-9 y^2=\) 36 which are perpendicular to the straight line \(5 x+2 y-10=0\) are

1 \(5(\mathrm{y}-3)=2\left(\mathrm{x}-\frac{\sqrt{117}}{2}\right)\)
2 \(2 x-5 y+10-2 \sqrt{18}=0\)
3 \(2 \mathrm{x}-5 \mathrm{y}-10-2 \sqrt{18}=0\)
4 None of the above
Jamia Millia Islamia-2009
Hyperbola

120804 The eccentricity of the hyperbola \(x^2 / 49-y^2 / 576\) \(=1\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: And,

1 \(25 / 24\)
2 \(25 / 7\)
3 \(1 / 25\)
4 \(49 / 576\)
And
Hyperbola

120805 Tangents are drawn to the hyperbola \(4 x^2-y^2=\) 36 at the points \(P\) and \(Q\). If these tangents intersect at the point \(T(0,3)\), then the area (in sq units) of \(\triangle \mathrm{PTQ}\) is

1 \(45 \sqrt{5}\)
2 \(54 \sqrt{3}\)
3 \(60 \sqrt{3}\)
4 \(36 \sqrt{5}\)
JEE Main-2018
For More Updates join with Us
Hyperbola

120801 A tangent to the curve \(9 b^2 x^2-4 a^2 y^2=36 a^2 b^2\) makes intercepts of unit length on each of the coordinate axes, then the point \((a, b)\) lies on

1 \(x^2-y^2=1\)
2 \(x^2+y^2=1\)
3 \(4 x^2-9 y^2=1\)
4 \(4 x^2+9 y^2=1\)
AP EAMCET-23.04.2018
Hyperbola

120802 The ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and the hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) have in common

1 centre only
2 centre, foci and directrices
3 centre, foci and vertices
4 centre and vertices only
Karnataka CET-2006
Hyperbola

120803 The equation of those tangents to \(4 x^2-9 y^2=\) 36 which are perpendicular to the straight line \(5 x+2 y-10=0\) are

1 \(5(\mathrm{y}-3)=2\left(\mathrm{x}-\frac{\sqrt{117}}{2}\right)\)
2 \(2 x-5 y+10-2 \sqrt{18}=0\)
3 \(2 \mathrm{x}-5 \mathrm{y}-10-2 \sqrt{18}=0\)
4 None of the above
Jamia Millia Islamia-2009
Hyperbola

120804 The eccentricity of the hyperbola \(x^2 / 49-y^2 / 576\) \(=1\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: And,

1 \(25 / 24\)
2 \(25 / 7\)
3 \(1 / 25\)
4 \(49 / 576\)
And
Hyperbola

120805 Tangents are drawn to the hyperbola \(4 x^2-y^2=\) 36 at the points \(P\) and \(Q\). If these tangents intersect at the point \(T(0,3)\), then the area (in sq units) of \(\triangle \mathrm{PTQ}\) is

1 \(45 \sqrt{5}\)
2 \(54 \sqrt{3}\)
3 \(60 \sqrt{3}\)
4 \(36 \sqrt{5}\)
JEE Main-2018
NEET DIGITAL LIBRARY
Hyperbola

120801 A tangent to the curve \(9 b^2 x^2-4 a^2 y^2=36 a^2 b^2\) makes intercepts of unit length on each of the coordinate axes, then the point \((a, b)\) lies on

1 \(x^2-y^2=1\)
2 \(x^2+y^2=1\)
3 \(4 x^2-9 y^2=1\)
4 \(4 x^2+9 y^2=1\)
AP EAMCET-23.04.2018
Hyperbola

120802 The ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and the hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) have in common

1 centre only
2 centre, foci and directrices
3 centre, foci and vertices
4 centre and vertices only
Karnataka CET-2006
Hyperbola

120803 The equation of those tangents to \(4 x^2-9 y^2=\) 36 which are perpendicular to the straight line \(5 x+2 y-10=0\) are

1 \(5(\mathrm{y}-3)=2\left(\mathrm{x}-\frac{\sqrt{117}}{2}\right)\)
2 \(2 x-5 y+10-2 \sqrt{18}=0\)
3 \(2 \mathrm{x}-5 \mathrm{y}-10-2 \sqrt{18}=0\)
4 None of the above
Jamia Millia Islamia-2009
Hyperbola

120804 The eccentricity of the hyperbola \(x^2 / 49-y^2 / 576\) \(=1\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: And,

1 \(25 / 24\)
2 \(25 / 7\)
3 \(1 / 25\)
4 \(49 / 576\)
And
Hyperbola

120805 Tangents are drawn to the hyperbola \(4 x^2-y^2=\) 36 at the points \(P\) and \(Q\). If these tangents intersect at the point \(T(0,3)\), then the area (in sq units) of \(\triangle \mathrm{PTQ}\) is

1 \(45 \sqrt{5}\)
2 \(54 \sqrt{3}\)
3 \(60 \sqrt{3}\)
4 \(36 \sqrt{5}\)
JEE Main-2018