120674
The eccentricity of the hyperbola \(3 x^2-4 y^2=-\) 12 is
1 \(\sqrt{\frac{7}{3}}\)
2 \(\frac{\sqrt{7}}{2}\)
3 \(-\frac{\sqrt{7}}{3}\)
4 \(-\frac{\sqrt{7}}{2}\)
Explanation:
A Given, \(3 x^2-4 y^2=-12\)
\(\frac{x^2}{4}-\frac{y^2}{3}=-1\)
\(\frac{y^2}{3}-\frac{x^2}{4}=1\)
Here, \(a=2, b=\sqrt{3}\)
Since, it is a conjugate hyperbola.
\(\therefore\) Eccentricity \((\mathrm{e})=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{4}{3}}=\sqrt{\frac{7}{3}}\)
JEEE-2010
Hyperbola
120675
The radius of the circle passing through the foci of the ellipse \(9 x^2+16 y^2=144\) and having its centre at \((0,3)\), is
1 4
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given,
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{9}=1\)
Here, \(\quad a^2=16, b^2=9\)
The eccentricity e of the ellipse is given by
\(\mathrm{e}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\)
So, the co-ordinate of the foci are \(=( \pm \mathrm{ae}, 0)=( \pm \sqrt{7}, 0)\)
\(\therefore\) Radius of the circle from the centre \((0,3)\) is-
\(=\sqrt{(\sqrt{7}-0)^2+(0-3)^2}\)
\(=\sqrt{7+9}=\sqrt{16}=4\)
Manipal UGET-2015
Hyperbola
120676
If \(5 x^2+\lambda y^2=20\) represent a rectangular
hyperbola then \(\lambda\) is equal to
1 5
2 4
3 -5
4 -4
Explanation:
C Given that, \(5 x^2+\lambda y^2=20\)
\(\frac{x^2}{4}+\frac{y^2}{20 / \lambda}=1\)
\(\frac{x^2}{4}-\frac{y^2}{(-20 / \lambda)}=1\)
This will represent a rectangular hyperbola,
\(\therefore \quad \mathrm{a}=\mathrm{b}\)
\(\Rightarrow \quad \frac{-20}{\lambda}=4 \Rightarrow \lambda=-5\)
JEEE-2011
Hyperbola
120677
The equation of the hyperbola whose directrix \(2 x+y=1\), focus \((1,1)\) and eccentricity \(\sqrt{3}\) is
1 \(7 x^2-2 y^2+12 x y+2 x-4 y+7=0\)
2 \(7 x^2+2 y^2+12 x y+2 x+4 y-7=0\)
3 \(7 x^2-2 y^2+12 x y-2 x+4 y-7=0\)
4 \(7 x^2-2 y^2+12 x y-2 x+4 y+7=0\)
Explanation:
C Given that,
\(\mathrm{e}=\sqrt{3}\)
Equation of directrix is \(2 \mathrm{x}+\mathrm{y}-1=0\)
Let, \(\mathrm{S}\) be the corresponding focus, then \(\mathrm{S}=(1,1)\)
Let, us consider \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the required hyperbola.
Let, PM be the length of the perpendicular from \(\mathrm{P}\) to directrix.
Then,
\(\frac{\mathrm{PS}}{\mathrm{PM}}=\mathrm{e}=\sqrt{3}\)
\(\Rightarrow \quad \mathrm{PS}=\sqrt{3} \mathrm{PM}\)
\(\Rightarrow \quad \mathrm{PS}^2=3 \mathrm{PM}^2\)
\((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=3\left|\frac{2 \mathrm{x}+\mathrm{y}-1}{\sqrt{5}}\right|^2\)
\(\Rightarrow 5\left(x^2+1-2 x+y^2+1-2 y\right)=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(\Rightarrow 5 x^2+5-10 x+5 y^2+5-10 y=12 x^2+3 y^2+3+12 x y-6 y-\)
\(12 \mathrm{x}\)
\(\Rightarrow 12 \mathrm{x}+7 \mathrm{x}^2-2 \mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-7=0\)
\(\Rightarrow 7 \mathrm{x}^2-2 \mathrm{y}^2+12 \mathrm{xy}-2 \mathrm{x}+4 \mathrm{y}-7=0\)
JEEE-2013
Hyperbola
120678
If the eccentricity of the hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is \(\frac{5}{4}\) and \(2 x+3 y-6=0\) is a focal chord of the hyperbola, then the length of transverse axis is equal to \(\qquad\)
1 \(\frac{24}{5}\)
2 \(\frac{5}{24}\)
3 \(\frac{12}{5}\)
4 \(\frac{6}{5}\)
Explanation:
A Given that,
Eccentricity \((\mathrm{e})=\frac{5}{4}\)
Equation of hyperbola, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\because\) The focus of hyperbola is \(( \pm\) ae, 0 )
\(2 x+3 y-6=0\) is the focal chord of the hyperbola for \((\mathrm{ae}, 0)\) [Considering one focus only]
\(\Rightarrow \quad 2 \mathrm{ae}+3(0)-6=0\)
\(\Rightarrow \quad \text { ae }=3\)
\(\Rightarrow \quad \mathrm{a}=\frac{3}{\mathrm{e}}\)
\(\mathrm{a}=\frac{3}{\frac{5}{4}} \quad\left[\because \mathrm{e}=\frac{5}{4}\right]\)
\(\Rightarrow \quad \mathrm{a}=\frac{12}{5}\)
Hence, the length of the transverse axis is \(2 \mathrm{a}=\frac{24}{5}\).
120674
The eccentricity of the hyperbola \(3 x^2-4 y^2=-\) 12 is
1 \(\sqrt{\frac{7}{3}}\)
2 \(\frac{\sqrt{7}}{2}\)
3 \(-\frac{\sqrt{7}}{3}\)
4 \(-\frac{\sqrt{7}}{2}\)
Explanation:
A Given, \(3 x^2-4 y^2=-12\)
\(\frac{x^2}{4}-\frac{y^2}{3}=-1\)
\(\frac{y^2}{3}-\frac{x^2}{4}=1\)
Here, \(a=2, b=\sqrt{3}\)
Since, it is a conjugate hyperbola.
\(\therefore\) Eccentricity \((\mathrm{e})=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{4}{3}}=\sqrt{\frac{7}{3}}\)
JEEE-2010
Hyperbola
120675
The radius of the circle passing through the foci of the ellipse \(9 x^2+16 y^2=144\) and having its centre at \((0,3)\), is
1 4
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given,
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{9}=1\)
Here, \(\quad a^2=16, b^2=9\)
The eccentricity e of the ellipse is given by
\(\mathrm{e}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\)
So, the co-ordinate of the foci are \(=( \pm \mathrm{ae}, 0)=( \pm \sqrt{7}, 0)\)
\(\therefore\) Radius of the circle from the centre \((0,3)\) is-
\(=\sqrt{(\sqrt{7}-0)^2+(0-3)^2}\)
\(=\sqrt{7+9}=\sqrt{16}=4\)
Manipal UGET-2015
Hyperbola
120676
If \(5 x^2+\lambda y^2=20\) represent a rectangular
hyperbola then \(\lambda\) is equal to
1 5
2 4
3 -5
4 -4
Explanation:
C Given that, \(5 x^2+\lambda y^2=20\)
\(\frac{x^2}{4}+\frac{y^2}{20 / \lambda}=1\)
\(\frac{x^2}{4}-\frac{y^2}{(-20 / \lambda)}=1\)
This will represent a rectangular hyperbola,
\(\therefore \quad \mathrm{a}=\mathrm{b}\)
\(\Rightarrow \quad \frac{-20}{\lambda}=4 \Rightarrow \lambda=-5\)
JEEE-2011
Hyperbola
120677
The equation of the hyperbola whose directrix \(2 x+y=1\), focus \((1,1)\) and eccentricity \(\sqrt{3}\) is
1 \(7 x^2-2 y^2+12 x y+2 x-4 y+7=0\)
2 \(7 x^2+2 y^2+12 x y+2 x+4 y-7=0\)
3 \(7 x^2-2 y^2+12 x y-2 x+4 y-7=0\)
4 \(7 x^2-2 y^2+12 x y-2 x+4 y+7=0\)
Explanation:
C Given that,
\(\mathrm{e}=\sqrt{3}\)
Equation of directrix is \(2 \mathrm{x}+\mathrm{y}-1=0\)
Let, \(\mathrm{S}\) be the corresponding focus, then \(\mathrm{S}=(1,1)\)
Let, us consider \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the required hyperbola.
Let, PM be the length of the perpendicular from \(\mathrm{P}\) to directrix.
Then,
\(\frac{\mathrm{PS}}{\mathrm{PM}}=\mathrm{e}=\sqrt{3}\)
\(\Rightarrow \quad \mathrm{PS}=\sqrt{3} \mathrm{PM}\)
\(\Rightarrow \quad \mathrm{PS}^2=3 \mathrm{PM}^2\)
\((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=3\left|\frac{2 \mathrm{x}+\mathrm{y}-1}{\sqrt{5}}\right|^2\)
\(\Rightarrow 5\left(x^2+1-2 x+y^2+1-2 y\right)=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(\Rightarrow 5 x^2+5-10 x+5 y^2+5-10 y=12 x^2+3 y^2+3+12 x y-6 y-\)
\(12 \mathrm{x}\)
\(\Rightarrow 12 \mathrm{x}+7 \mathrm{x}^2-2 \mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-7=0\)
\(\Rightarrow 7 \mathrm{x}^2-2 \mathrm{y}^2+12 \mathrm{xy}-2 \mathrm{x}+4 \mathrm{y}-7=0\)
JEEE-2013
Hyperbola
120678
If the eccentricity of the hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is \(\frac{5}{4}\) and \(2 x+3 y-6=0\) is a focal chord of the hyperbola, then the length of transverse axis is equal to \(\qquad\)
1 \(\frac{24}{5}\)
2 \(\frac{5}{24}\)
3 \(\frac{12}{5}\)
4 \(\frac{6}{5}\)
Explanation:
A Given that,
Eccentricity \((\mathrm{e})=\frac{5}{4}\)
Equation of hyperbola, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\because\) The focus of hyperbola is \(( \pm\) ae, 0 )
\(2 x+3 y-6=0\) is the focal chord of the hyperbola for \((\mathrm{ae}, 0)\) [Considering one focus only]
\(\Rightarrow \quad 2 \mathrm{ae}+3(0)-6=0\)
\(\Rightarrow \quad \text { ae }=3\)
\(\Rightarrow \quad \mathrm{a}=\frac{3}{\mathrm{e}}\)
\(\mathrm{a}=\frac{3}{\frac{5}{4}} \quad\left[\because \mathrm{e}=\frac{5}{4}\right]\)
\(\Rightarrow \quad \mathrm{a}=\frac{12}{5}\)
Hence, the length of the transverse axis is \(2 \mathrm{a}=\frac{24}{5}\).
120674
The eccentricity of the hyperbola \(3 x^2-4 y^2=-\) 12 is
1 \(\sqrt{\frac{7}{3}}\)
2 \(\frac{\sqrt{7}}{2}\)
3 \(-\frac{\sqrt{7}}{3}\)
4 \(-\frac{\sqrt{7}}{2}\)
Explanation:
A Given, \(3 x^2-4 y^2=-12\)
\(\frac{x^2}{4}-\frac{y^2}{3}=-1\)
\(\frac{y^2}{3}-\frac{x^2}{4}=1\)
Here, \(a=2, b=\sqrt{3}\)
Since, it is a conjugate hyperbola.
\(\therefore\) Eccentricity \((\mathrm{e})=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{4}{3}}=\sqrt{\frac{7}{3}}\)
JEEE-2010
Hyperbola
120675
The radius of the circle passing through the foci of the ellipse \(9 x^2+16 y^2=144\) and having its centre at \((0,3)\), is
1 4
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given,
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{9}=1\)
Here, \(\quad a^2=16, b^2=9\)
The eccentricity e of the ellipse is given by
\(\mathrm{e}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\)
So, the co-ordinate of the foci are \(=( \pm \mathrm{ae}, 0)=( \pm \sqrt{7}, 0)\)
\(\therefore\) Radius of the circle from the centre \((0,3)\) is-
\(=\sqrt{(\sqrt{7}-0)^2+(0-3)^2}\)
\(=\sqrt{7+9}=\sqrt{16}=4\)
Manipal UGET-2015
Hyperbola
120676
If \(5 x^2+\lambda y^2=20\) represent a rectangular
hyperbola then \(\lambda\) is equal to
1 5
2 4
3 -5
4 -4
Explanation:
C Given that, \(5 x^2+\lambda y^2=20\)
\(\frac{x^2}{4}+\frac{y^2}{20 / \lambda}=1\)
\(\frac{x^2}{4}-\frac{y^2}{(-20 / \lambda)}=1\)
This will represent a rectangular hyperbola,
\(\therefore \quad \mathrm{a}=\mathrm{b}\)
\(\Rightarrow \quad \frac{-20}{\lambda}=4 \Rightarrow \lambda=-5\)
JEEE-2011
Hyperbola
120677
The equation of the hyperbola whose directrix \(2 x+y=1\), focus \((1,1)\) and eccentricity \(\sqrt{3}\) is
1 \(7 x^2-2 y^2+12 x y+2 x-4 y+7=0\)
2 \(7 x^2+2 y^2+12 x y+2 x+4 y-7=0\)
3 \(7 x^2-2 y^2+12 x y-2 x+4 y-7=0\)
4 \(7 x^2-2 y^2+12 x y-2 x+4 y+7=0\)
Explanation:
C Given that,
\(\mathrm{e}=\sqrt{3}\)
Equation of directrix is \(2 \mathrm{x}+\mathrm{y}-1=0\)
Let, \(\mathrm{S}\) be the corresponding focus, then \(\mathrm{S}=(1,1)\)
Let, us consider \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the required hyperbola.
Let, PM be the length of the perpendicular from \(\mathrm{P}\) to directrix.
Then,
\(\frac{\mathrm{PS}}{\mathrm{PM}}=\mathrm{e}=\sqrt{3}\)
\(\Rightarrow \quad \mathrm{PS}=\sqrt{3} \mathrm{PM}\)
\(\Rightarrow \quad \mathrm{PS}^2=3 \mathrm{PM}^2\)
\((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=3\left|\frac{2 \mathrm{x}+\mathrm{y}-1}{\sqrt{5}}\right|^2\)
\(\Rightarrow 5\left(x^2+1-2 x+y^2+1-2 y\right)=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(\Rightarrow 5 x^2+5-10 x+5 y^2+5-10 y=12 x^2+3 y^2+3+12 x y-6 y-\)
\(12 \mathrm{x}\)
\(\Rightarrow 12 \mathrm{x}+7 \mathrm{x}^2-2 \mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-7=0\)
\(\Rightarrow 7 \mathrm{x}^2-2 \mathrm{y}^2+12 \mathrm{xy}-2 \mathrm{x}+4 \mathrm{y}-7=0\)
JEEE-2013
Hyperbola
120678
If the eccentricity of the hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is \(\frac{5}{4}\) and \(2 x+3 y-6=0\) is a focal chord of the hyperbola, then the length of transverse axis is equal to \(\qquad\)
1 \(\frac{24}{5}\)
2 \(\frac{5}{24}\)
3 \(\frac{12}{5}\)
4 \(\frac{6}{5}\)
Explanation:
A Given that,
Eccentricity \((\mathrm{e})=\frac{5}{4}\)
Equation of hyperbola, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\because\) The focus of hyperbola is \(( \pm\) ae, 0 )
\(2 x+3 y-6=0\) is the focal chord of the hyperbola for \((\mathrm{ae}, 0)\) [Considering one focus only]
\(\Rightarrow \quad 2 \mathrm{ae}+3(0)-6=0\)
\(\Rightarrow \quad \text { ae }=3\)
\(\Rightarrow \quad \mathrm{a}=\frac{3}{\mathrm{e}}\)
\(\mathrm{a}=\frac{3}{\frac{5}{4}} \quad\left[\because \mathrm{e}=\frac{5}{4}\right]\)
\(\Rightarrow \quad \mathrm{a}=\frac{12}{5}\)
Hence, the length of the transverse axis is \(2 \mathrm{a}=\frac{24}{5}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120674
The eccentricity of the hyperbola \(3 x^2-4 y^2=-\) 12 is
1 \(\sqrt{\frac{7}{3}}\)
2 \(\frac{\sqrt{7}}{2}\)
3 \(-\frac{\sqrt{7}}{3}\)
4 \(-\frac{\sqrt{7}}{2}\)
Explanation:
A Given, \(3 x^2-4 y^2=-12\)
\(\frac{x^2}{4}-\frac{y^2}{3}=-1\)
\(\frac{y^2}{3}-\frac{x^2}{4}=1\)
Here, \(a=2, b=\sqrt{3}\)
Since, it is a conjugate hyperbola.
\(\therefore\) Eccentricity \((\mathrm{e})=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{4}{3}}=\sqrt{\frac{7}{3}}\)
JEEE-2010
Hyperbola
120675
The radius of the circle passing through the foci of the ellipse \(9 x^2+16 y^2=144\) and having its centre at \((0,3)\), is
1 4
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given,
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{9}=1\)
Here, \(\quad a^2=16, b^2=9\)
The eccentricity e of the ellipse is given by
\(\mathrm{e}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\)
So, the co-ordinate of the foci are \(=( \pm \mathrm{ae}, 0)=( \pm \sqrt{7}, 0)\)
\(\therefore\) Radius of the circle from the centre \((0,3)\) is-
\(=\sqrt{(\sqrt{7}-0)^2+(0-3)^2}\)
\(=\sqrt{7+9}=\sqrt{16}=4\)
Manipal UGET-2015
Hyperbola
120676
If \(5 x^2+\lambda y^2=20\) represent a rectangular
hyperbola then \(\lambda\) is equal to
1 5
2 4
3 -5
4 -4
Explanation:
C Given that, \(5 x^2+\lambda y^2=20\)
\(\frac{x^2}{4}+\frac{y^2}{20 / \lambda}=1\)
\(\frac{x^2}{4}-\frac{y^2}{(-20 / \lambda)}=1\)
This will represent a rectangular hyperbola,
\(\therefore \quad \mathrm{a}=\mathrm{b}\)
\(\Rightarrow \quad \frac{-20}{\lambda}=4 \Rightarrow \lambda=-5\)
JEEE-2011
Hyperbola
120677
The equation of the hyperbola whose directrix \(2 x+y=1\), focus \((1,1)\) and eccentricity \(\sqrt{3}\) is
1 \(7 x^2-2 y^2+12 x y+2 x-4 y+7=0\)
2 \(7 x^2+2 y^2+12 x y+2 x+4 y-7=0\)
3 \(7 x^2-2 y^2+12 x y-2 x+4 y-7=0\)
4 \(7 x^2-2 y^2+12 x y-2 x+4 y+7=0\)
Explanation:
C Given that,
\(\mathrm{e}=\sqrt{3}\)
Equation of directrix is \(2 \mathrm{x}+\mathrm{y}-1=0\)
Let, \(\mathrm{S}\) be the corresponding focus, then \(\mathrm{S}=(1,1)\)
Let, us consider \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the required hyperbola.
Let, PM be the length of the perpendicular from \(\mathrm{P}\) to directrix.
Then,
\(\frac{\mathrm{PS}}{\mathrm{PM}}=\mathrm{e}=\sqrt{3}\)
\(\Rightarrow \quad \mathrm{PS}=\sqrt{3} \mathrm{PM}\)
\(\Rightarrow \quad \mathrm{PS}^2=3 \mathrm{PM}^2\)
\((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=3\left|\frac{2 \mathrm{x}+\mathrm{y}-1}{\sqrt{5}}\right|^2\)
\(\Rightarrow 5\left(x^2+1-2 x+y^2+1-2 y\right)=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(\Rightarrow 5 x^2+5-10 x+5 y^2+5-10 y=12 x^2+3 y^2+3+12 x y-6 y-\)
\(12 \mathrm{x}\)
\(\Rightarrow 12 \mathrm{x}+7 \mathrm{x}^2-2 \mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-7=0\)
\(\Rightarrow 7 \mathrm{x}^2-2 \mathrm{y}^2+12 \mathrm{xy}-2 \mathrm{x}+4 \mathrm{y}-7=0\)
JEEE-2013
Hyperbola
120678
If the eccentricity of the hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is \(\frac{5}{4}\) and \(2 x+3 y-6=0\) is a focal chord of the hyperbola, then the length of transverse axis is equal to \(\qquad\)
1 \(\frac{24}{5}\)
2 \(\frac{5}{24}\)
3 \(\frac{12}{5}\)
4 \(\frac{6}{5}\)
Explanation:
A Given that,
Eccentricity \((\mathrm{e})=\frac{5}{4}\)
Equation of hyperbola, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\because\) The focus of hyperbola is \(( \pm\) ae, 0 )
\(2 x+3 y-6=0\) is the focal chord of the hyperbola for \((\mathrm{ae}, 0)\) [Considering one focus only]
\(\Rightarrow \quad 2 \mathrm{ae}+3(0)-6=0\)
\(\Rightarrow \quad \text { ae }=3\)
\(\Rightarrow \quad \mathrm{a}=\frac{3}{\mathrm{e}}\)
\(\mathrm{a}=\frac{3}{\frac{5}{4}} \quad\left[\because \mathrm{e}=\frac{5}{4}\right]\)
\(\Rightarrow \quad \mathrm{a}=\frac{12}{5}\)
Hence, the length of the transverse axis is \(2 \mathrm{a}=\frac{24}{5}\).
120674
The eccentricity of the hyperbola \(3 x^2-4 y^2=-\) 12 is
1 \(\sqrt{\frac{7}{3}}\)
2 \(\frac{\sqrt{7}}{2}\)
3 \(-\frac{\sqrt{7}}{3}\)
4 \(-\frac{\sqrt{7}}{2}\)
Explanation:
A Given, \(3 x^2-4 y^2=-12\)
\(\frac{x^2}{4}-\frac{y^2}{3}=-1\)
\(\frac{y^2}{3}-\frac{x^2}{4}=1\)
Here, \(a=2, b=\sqrt{3}\)
Since, it is a conjugate hyperbola.
\(\therefore\) Eccentricity \((\mathrm{e})=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{4}{3}}=\sqrt{\frac{7}{3}}\)
JEEE-2010
Hyperbola
120675
The radius of the circle passing through the foci of the ellipse \(9 x^2+16 y^2=144\) and having its centre at \((0,3)\), is
1 4
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given,
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{9}=1\)
Here, \(\quad a^2=16, b^2=9\)
The eccentricity e of the ellipse is given by
\(\mathrm{e}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\)
So, the co-ordinate of the foci are \(=( \pm \mathrm{ae}, 0)=( \pm \sqrt{7}, 0)\)
\(\therefore\) Radius of the circle from the centre \((0,3)\) is-
\(=\sqrt{(\sqrt{7}-0)^2+(0-3)^2}\)
\(=\sqrt{7+9}=\sqrt{16}=4\)
Manipal UGET-2015
Hyperbola
120676
If \(5 x^2+\lambda y^2=20\) represent a rectangular
hyperbola then \(\lambda\) is equal to
1 5
2 4
3 -5
4 -4
Explanation:
C Given that, \(5 x^2+\lambda y^2=20\)
\(\frac{x^2}{4}+\frac{y^2}{20 / \lambda}=1\)
\(\frac{x^2}{4}-\frac{y^2}{(-20 / \lambda)}=1\)
This will represent a rectangular hyperbola,
\(\therefore \quad \mathrm{a}=\mathrm{b}\)
\(\Rightarrow \quad \frac{-20}{\lambda}=4 \Rightarrow \lambda=-5\)
JEEE-2011
Hyperbola
120677
The equation of the hyperbola whose directrix \(2 x+y=1\), focus \((1,1)\) and eccentricity \(\sqrt{3}\) is
1 \(7 x^2-2 y^2+12 x y+2 x-4 y+7=0\)
2 \(7 x^2+2 y^2+12 x y+2 x+4 y-7=0\)
3 \(7 x^2-2 y^2+12 x y-2 x+4 y-7=0\)
4 \(7 x^2-2 y^2+12 x y-2 x+4 y+7=0\)
Explanation:
C Given that,
\(\mathrm{e}=\sqrt{3}\)
Equation of directrix is \(2 \mathrm{x}+\mathrm{y}-1=0\)
Let, \(\mathrm{S}\) be the corresponding focus, then \(\mathrm{S}=(1,1)\)
Let, us consider \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the required hyperbola.
Let, PM be the length of the perpendicular from \(\mathrm{P}\) to directrix.
Then,
\(\frac{\mathrm{PS}}{\mathrm{PM}}=\mathrm{e}=\sqrt{3}\)
\(\Rightarrow \quad \mathrm{PS}=\sqrt{3} \mathrm{PM}\)
\(\Rightarrow \quad \mathrm{PS}^2=3 \mathrm{PM}^2\)
\((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=3\left|\frac{2 \mathrm{x}+\mathrm{y}-1}{\sqrt{5}}\right|^2\)
\(\Rightarrow 5\left(x^2+1-2 x+y^2+1-2 y\right)=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(\Rightarrow 5 x^2+5-10 x+5 y^2+5-10 y=12 x^2+3 y^2+3+12 x y-6 y-\)
\(12 \mathrm{x}\)
\(\Rightarrow 12 \mathrm{x}+7 \mathrm{x}^2-2 \mathrm{y}^2-2 \mathrm{x}+4 \mathrm{y}-7=0\)
\(\Rightarrow 7 \mathrm{x}^2-2 \mathrm{y}^2+12 \mathrm{xy}-2 \mathrm{x}+4 \mathrm{y}-7=0\)
JEEE-2013
Hyperbola
120678
If the eccentricity of the hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is \(\frac{5}{4}\) and \(2 x+3 y-6=0\) is a focal chord of the hyperbola, then the length of transverse axis is equal to \(\qquad\)
1 \(\frac{24}{5}\)
2 \(\frac{5}{24}\)
3 \(\frac{12}{5}\)
4 \(\frac{6}{5}\)
Explanation:
A Given that,
Eccentricity \((\mathrm{e})=\frac{5}{4}\)
Equation of hyperbola, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\because\) The focus of hyperbola is \(( \pm\) ae, 0 )
\(2 x+3 y-6=0\) is the focal chord of the hyperbola for \((\mathrm{ae}, 0)\) [Considering one focus only]
\(\Rightarrow \quad 2 \mathrm{ae}+3(0)-6=0\)
\(\Rightarrow \quad \text { ae }=3\)
\(\Rightarrow \quad \mathrm{a}=\frac{3}{\mathrm{e}}\)
\(\mathrm{a}=\frac{3}{\frac{5}{4}} \quad\left[\because \mathrm{e}=\frac{5}{4}\right]\)
\(\Rightarrow \quad \mathrm{a}=\frac{12}{5}\)
Hence, the length of the transverse axis is \(2 \mathrm{a}=\frac{24}{5}\).