120701
The eccentricity of the hyperbola \(5 x^2-4 y^2+20 x+8 y=4\) is
1 \(\sqrt{2}\)
2 \(\frac{3}{2}\)
3 2
4 3
Explanation:
B Given hyperbola is,
\(5 x^2-4 y^2+20 x+8 y=4\)
\(5\left(x^2+4 x+4\right)-20-4\left(y^2-2 y+1\right)+4=4\)
\(5\left(x^2+2\right)^2-4(y-1)^2=20\)
Divide both side by ' 20 ',
\(\frac{5(x+2)^2}{20}-\frac{4(y-1)^2}{20}=\frac{20}{20}\)
\(\frac{(x+2)^2}{4}-\frac{(y-1)^2}{5}=1 \quad\left[\because a^2=4 \text { and } b^2=5\right]\)
we know that,
\(\text { Eccentricity }(\mathrm{e}) =\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{\frac{4+5}{4}}\)
\(\mathrm{e} =\sqrt{\frac{9}{4}}=\frac{3}{2}\)
BCECE-2016
Hyperbola
120702
The eccentricity of the hyperbola
\(\frac{\sqrt{1999}}{3}\left(\mathrm{x}^2-\mathrm{y}^2\right)=1 \text { is }\)
1 \(\sqrt{2}\)
2 2
3 \(2 \sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
A : Given equation of hyperbola,
\(\frac{\sqrt{1999}}{3}\left(x^2-y^2\right)=1\)
So, equation of hyperbola is \(\frac{x^2}{3 / \sqrt{1999}}-\frac{y^2}{3 / \sqrt{1999}}=1\)
Here \(\mathrm{a}^2=\mathrm{b}^2=\frac{3}{\sqrt{1999}}\)
\(\therefore \text { Eccentricity }(\mathrm{e})=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}\)
BCECE-2015
Hyperbola
120703
If \(t\) is a parameter, then \(x=a\left(t+\frac{1}{t}\right)\) and \(y=b\left(t-\frac{1}{t}\right)\) represent
1 an ellipse
2 a circle
3 a pair of straight lines
4 a hyperbola
Explanation:
D Given equation,
\(x=a\left(t+\frac{1}{t}\right) \text { and } y=b\left(t-\frac{1}{t}\right)\)
\(\frac{x}{a}=\left(t+\frac{1}{t}\right) \text { and } \frac{y}{b}=\left(t-\frac{1}{t}\right)\)
On squaring and subtracting, we get-
\(\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2=\left(t+\frac{1}{t}\right)^2-\left(t-\frac{1}{t}\right)^2\)
\(=\left(t^2+\frac{1}{t^2}+2\right)-\left(t^2+\frac{1}{t^2}-2\right)\)
\(=t^2+\frac{1}{t^2}+2-t^2-\frac{1}{t^2}+2 \Rightarrow 4\)
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=4\)
\(\frac{x^2}{(2 a)^2}-\frac{y^2}{(2 b)^2}=1\)Which represents a hyperbola.
BCECE-2014
Hyperbola
120705
If focii of \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) coincide with the foci of \(\frac{x^2}{25}+\frac{y^2}{9}=1\) and eccentricity of the hyperbola is 2 , then
1 \(a^2+b^2=16\)
2 there is a director circle of the hyperbola
3 \((\mathrm{a}, \mathrm{b}, \mathrm{d})\)
4 length of latusrectum of the hyperbola is 12
Explanation:
A Given that, \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
Which is an ellipse
\(\because \quad \mathrm{a}^2=25, \mathrm{~b}^2=9\)
\(\mathrm{a}=5, \quad \mathrm{~b}=3\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{9}{25}} \quad \mathrm{e}=\sqrt{\frac{16}{25}} \quad \mathrm{e}=\frac{4}{5}\)
Hence, the foci are \((-4,0)\) and \((4,0)\)
For the hyperbola,
\(\mathrm{ae}=4, \quad \mathrm{e}=2\)
\(\mathrm{a}=2\)
\(\mathrm{~b}^2=4(4-1)=12\)
\(\mathrm{~b}=\sqrt{12}\)
\(\mathrm{a}^2+\mathrm{b}^2=2^2+12=16\)
Since \(b^2>a^2\), so there is no director circle to the hyperbola.
Length of latus rectum \(=\frac{2 b^2}{a}=\frac{12 \times 2}{2}\)
\(=12\)
120701
The eccentricity of the hyperbola \(5 x^2-4 y^2+20 x+8 y=4\) is
1 \(\sqrt{2}\)
2 \(\frac{3}{2}\)
3 2
4 3
Explanation:
B Given hyperbola is,
\(5 x^2-4 y^2+20 x+8 y=4\)
\(5\left(x^2+4 x+4\right)-20-4\left(y^2-2 y+1\right)+4=4\)
\(5\left(x^2+2\right)^2-4(y-1)^2=20\)
Divide both side by ' 20 ',
\(\frac{5(x+2)^2}{20}-\frac{4(y-1)^2}{20}=\frac{20}{20}\)
\(\frac{(x+2)^2}{4}-\frac{(y-1)^2}{5}=1 \quad\left[\because a^2=4 \text { and } b^2=5\right]\)
we know that,
\(\text { Eccentricity }(\mathrm{e}) =\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{\frac{4+5}{4}}\)
\(\mathrm{e} =\sqrt{\frac{9}{4}}=\frac{3}{2}\)
BCECE-2016
Hyperbola
120702
The eccentricity of the hyperbola
\(\frac{\sqrt{1999}}{3}\left(\mathrm{x}^2-\mathrm{y}^2\right)=1 \text { is }\)
1 \(\sqrt{2}\)
2 2
3 \(2 \sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
A : Given equation of hyperbola,
\(\frac{\sqrt{1999}}{3}\left(x^2-y^2\right)=1\)
So, equation of hyperbola is \(\frac{x^2}{3 / \sqrt{1999}}-\frac{y^2}{3 / \sqrt{1999}}=1\)
Here \(\mathrm{a}^2=\mathrm{b}^2=\frac{3}{\sqrt{1999}}\)
\(\therefore \text { Eccentricity }(\mathrm{e})=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}\)
BCECE-2015
Hyperbola
120703
If \(t\) is a parameter, then \(x=a\left(t+\frac{1}{t}\right)\) and \(y=b\left(t-\frac{1}{t}\right)\) represent
1 an ellipse
2 a circle
3 a pair of straight lines
4 a hyperbola
Explanation:
D Given equation,
\(x=a\left(t+\frac{1}{t}\right) \text { and } y=b\left(t-\frac{1}{t}\right)\)
\(\frac{x}{a}=\left(t+\frac{1}{t}\right) \text { and } \frac{y}{b}=\left(t-\frac{1}{t}\right)\)
On squaring and subtracting, we get-
\(\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2=\left(t+\frac{1}{t}\right)^2-\left(t-\frac{1}{t}\right)^2\)
\(=\left(t^2+\frac{1}{t^2}+2\right)-\left(t^2+\frac{1}{t^2}-2\right)\)
\(=t^2+\frac{1}{t^2}+2-t^2-\frac{1}{t^2}+2 \Rightarrow 4\)
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=4\)
\(\frac{x^2}{(2 a)^2}-\frac{y^2}{(2 b)^2}=1\)Which represents a hyperbola.
BCECE-2014
Hyperbola
120705
If focii of \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) coincide with the foci of \(\frac{x^2}{25}+\frac{y^2}{9}=1\) and eccentricity of the hyperbola is 2 , then
1 \(a^2+b^2=16\)
2 there is a director circle of the hyperbola
3 \((\mathrm{a}, \mathrm{b}, \mathrm{d})\)
4 length of latusrectum of the hyperbola is 12
Explanation:
A Given that, \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
Which is an ellipse
\(\because \quad \mathrm{a}^2=25, \mathrm{~b}^2=9\)
\(\mathrm{a}=5, \quad \mathrm{~b}=3\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{9}{25}} \quad \mathrm{e}=\sqrt{\frac{16}{25}} \quad \mathrm{e}=\frac{4}{5}\)
Hence, the foci are \((-4,0)\) and \((4,0)\)
For the hyperbola,
\(\mathrm{ae}=4, \quad \mathrm{e}=2\)
\(\mathrm{a}=2\)
\(\mathrm{~b}^2=4(4-1)=12\)
\(\mathrm{~b}=\sqrt{12}\)
\(\mathrm{a}^2+\mathrm{b}^2=2^2+12=16\)
Since \(b^2>a^2\), so there is no director circle to the hyperbola.
Length of latus rectum \(=\frac{2 b^2}{a}=\frac{12 \times 2}{2}\)
\(=12\)
120701
The eccentricity of the hyperbola \(5 x^2-4 y^2+20 x+8 y=4\) is
1 \(\sqrt{2}\)
2 \(\frac{3}{2}\)
3 2
4 3
Explanation:
B Given hyperbola is,
\(5 x^2-4 y^2+20 x+8 y=4\)
\(5\left(x^2+4 x+4\right)-20-4\left(y^2-2 y+1\right)+4=4\)
\(5\left(x^2+2\right)^2-4(y-1)^2=20\)
Divide both side by ' 20 ',
\(\frac{5(x+2)^2}{20}-\frac{4(y-1)^2}{20}=\frac{20}{20}\)
\(\frac{(x+2)^2}{4}-\frac{(y-1)^2}{5}=1 \quad\left[\because a^2=4 \text { and } b^2=5\right]\)
we know that,
\(\text { Eccentricity }(\mathrm{e}) =\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{\frac{4+5}{4}}\)
\(\mathrm{e} =\sqrt{\frac{9}{4}}=\frac{3}{2}\)
BCECE-2016
Hyperbola
120702
The eccentricity of the hyperbola
\(\frac{\sqrt{1999}}{3}\left(\mathrm{x}^2-\mathrm{y}^2\right)=1 \text { is }\)
1 \(\sqrt{2}\)
2 2
3 \(2 \sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
A : Given equation of hyperbola,
\(\frac{\sqrt{1999}}{3}\left(x^2-y^2\right)=1\)
So, equation of hyperbola is \(\frac{x^2}{3 / \sqrt{1999}}-\frac{y^2}{3 / \sqrt{1999}}=1\)
Here \(\mathrm{a}^2=\mathrm{b}^2=\frac{3}{\sqrt{1999}}\)
\(\therefore \text { Eccentricity }(\mathrm{e})=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}\)
BCECE-2015
Hyperbola
120703
If \(t\) is a parameter, then \(x=a\left(t+\frac{1}{t}\right)\) and \(y=b\left(t-\frac{1}{t}\right)\) represent
1 an ellipse
2 a circle
3 a pair of straight lines
4 a hyperbola
Explanation:
D Given equation,
\(x=a\left(t+\frac{1}{t}\right) \text { and } y=b\left(t-\frac{1}{t}\right)\)
\(\frac{x}{a}=\left(t+\frac{1}{t}\right) \text { and } \frac{y}{b}=\left(t-\frac{1}{t}\right)\)
On squaring and subtracting, we get-
\(\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2=\left(t+\frac{1}{t}\right)^2-\left(t-\frac{1}{t}\right)^2\)
\(=\left(t^2+\frac{1}{t^2}+2\right)-\left(t^2+\frac{1}{t^2}-2\right)\)
\(=t^2+\frac{1}{t^2}+2-t^2-\frac{1}{t^2}+2 \Rightarrow 4\)
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=4\)
\(\frac{x^2}{(2 a)^2}-\frac{y^2}{(2 b)^2}=1\)Which represents a hyperbola.
BCECE-2014
Hyperbola
120705
If focii of \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) coincide with the foci of \(\frac{x^2}{25}+\frac{y^2}{9}=1\) and eccentricity of the hyperbola is 2 , then
1 \(a^2+b^2=16\)
2 there is a director circle of the hyperbola
3 \((\mathrm{a}, \mathrm{b}, \mathrm{d})\)
4 length of latusrectum of the hyperbola is 12
Explanation:
A Given that, \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
Which is an ellipse
\(\because \quad \mathrm{a}^2=25, \mathrm{~b}^2=9\)
\(\mathrm{a}=5, \quad \mathrm{~b}=3\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{9}{25}} \quad \mathrm{e}=\sqrt{\frac{16}{25}} \quad \mathrm{e}=\frac{4}{5}\)
Hence, the foci are \((-4,0)\) and \((4,0)\)
For the hyperbola,
\(\mathrm{ae}=4, \quad \mathrm{e}=2\)
\(\mathrm{a}=2\)
\(\mathrm{~b}^2=4(4-1)=12\)
\(\mathrm{~b}=\sqrt{12}\)
\(\mathrm{a}^2+\mathrm{b}^2=2^2+12=16\)
Since \(b^2>a^2\), so there is no director circle to the hyperbola.
Length of latus rectum \(=\frac{2 b^2}{a}=\frac{12 \times 2}{2}\)
\(=12\)
120701
The eccentricity of the hyperbola \(5 x^2-4 y^2+20 x+8 y=4\) is
1 \(\sqrt{2}\)
2 \(\frac{3}{2}\)
3 2
4 3
Explanation:
B Given hyperbola is,
\(5 x^2-4 y^2+20 x+8 y=4\)
\(5\left(x^2+4 x+4\right)-20-4\left(y^2-2 y+1\right)+4=4\)
\(5\left(x^2+2\right)^2-4(y-1)^2=20\)
Divide both side by ' 20 ',
\(\frac{5(x+2)^2}{20}-\frac{4(y-1)^2}{20}=\frac{20}{20}\)
\(\frac{(x+2)^2}{4}-\frac{(y-1)^2}{5}=1 \quad\left[\because a^2=4 \text { and } b^2=5\right]\)
we know that,
\(\text { Eccentricity }(\mathrm{e}) =\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{\frac{4+5}{4}}\)
\(\mathrm{e} =\sqrt{\frac{9}{4}}=\frac{3}{2}\)
BCECE-2016
Hyperbola
120702
The eccentricity of the hyperbola
\(\frac{\sqrt{1999}}{3}\left(\mathrm{x}^2-\mathrm{y}^2\right)=1 \text { is }\)
1 \(\sqrt{2}\)
2 2
3 \(2 \sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
A : Given equation of hyperbola,
\(\frac{\sqrt{1999}}{3}\left(x^2-y^2\right)=1\)
So, equation of hyperbola is \(\frac{x^2}{3 / \sqrt{1999}}-\frac{y^2}{3 / \sqrt{1999}}=1\)
Here \(\mathrm{a}^2=\mathrm{b}^2=\frac{3}{\sqrt{1999}}\)
\(\therefore \text { Eccentricity }(\mathrm{e})=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}\)
BCECE-2015
Hyperbola
120703
If \(t\) is a parameter, then \(x=a\left(t+\frac{1}{t}\right)\) and \(y=b\left(t-\frac{1}{t}\right)\) represent
1 an ellipse
2 a circle
3 a pair of straight lines
4 a hyperbola
Explanation:
D Given equation,
\(x=a\left(t+\frac{1}{t}\right) \text { and } y=b\left(t-\frac{1}{t}\right)\)
\(\frac{x}{a}=\left(t+\frac{1}{t}\right) \text { and } \frac{y}{b}=\left(t-\frac{1}{t}\right)\)
On squaring and subtracting, we get-
\(\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2=\left(t+\frac{1}{t}\right)^2-\left(t-\frac{1}{t}\right)^2\)
\(=\left(t^2+\frac{1}{t^2}+2\right)-\left(t^2+\frac{1}{t^2}-2\right)\)
\(=t^2+\frac{1}{t^2}+2-t^2-\frac{1}{t^2}+2 \Rightarrow 4\)
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=4\)
\(\frac{x^2}{(2 a)^2}-\frac{y^2}{(2 b)^2}=1\)Which represents a hyperbola.
BCECE-2014
Hyperbola
120705
If focii of \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) coincide with the foci of \(\frac{x^2}{25}+\frac{y^2}{9}=1\) and eccentricity of the hyperbola is 2 , then
1 \(a^2+b^2=16\)
2 there is a director circle of the hyperbola
3 \((\mathrm{a}, \mathrm{b}, \mathrm{d})\)
4 length of latusrectum of the hyperbola is 12
Explanation:
A Given that, \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
Which is an ellipse
\(\because \quad \mathrm{a}^2=25, \mathrm{~b}^2=9\)
\(\mathrm{a}=5, \quad \mathrm{~b}=3\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{9}{25}} \quad \mathrm{e}=\sqrt{\frac{16}{25}} \quad \mathrm{e}=\frac{4}{5}\)
Hence, the foci are \((-4,0)\) and \((4,0)\)
For the hyperbola,
\(\mathrm{ae}=4, \quad \mathrm{e}=2\)
\(\mathrm{a}=2\)
\(\mathrm{~b}^2=4(4-1)=12\)
\(\mathrm{~b}=\sqrt{12}\)
\(\mathrm{a}^2+\mathrm{b}^2=2^2+12=16\)
Since \(b^2>a^2\), so there is no director circle to the hyperbola.
Length of latus rectum \(=\frac{2 b^2}{a}=\frac{12 \times 2}{2}\)
\(=12\)
