120680
If the eccentricity of a hyperbola is \(5 / 3\), then the eccentricity of its conjugate is
1 \(5 / 3\)
2 \(5 / 4\)
3 5
4 non existent
Explanation:
B Given that, Eccentricity of hyperbola is \(\frac{5}{3}\)
Let \(e_1=\) eccentricity of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
And \(e_2=\) eccentricity of its conjugate \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\)
\(\mathrm{e}_1=\sqrt{\frac{\mathrm{b}^2}{\mathrm{a}^2}+1} \text { and } \mathrm{e}_2=\sqrt{\frac{\mathrm{a}^2}{\mathrm{~b}^2}+1}\)
\(\because \mathrm{e}_1=\frac{5}{3}\)
\(\therefore \frac{25}{9}=\frac{\mathrm{b}^2}{\mathrm{a}^2}+1 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{25}{9}-1=\frac{16}{9}\)
\(\therefore \mathrm{e}_2=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
COMEDK-2011
Hyperbola
120681
The equation of auxiliary circle of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{9}=1\) is
1 \(x^2+y^2=4\)
2 \(x^2+y^2=9\)
3 \(x^2+y^2=13\)
4 \(x^2+y^2=5\)
Explanation:
A Given, hyperbola is \(\frac{\mathrm{x}^2}{4}-\frac{\mathrm{y}^2}{9}=1\)
Auxiliary circle is the circle between the two curves of hyperbola.
\(\therefore(\mathrm{a}, 0)\) and \((-\mathrm{a}, 0)\) touches the circle with centre \((0,0)\)
\(\therefore\) Equation of circle is
\((x-0)^2+(y-0)^2=\left(\sqrt{(a-0)^2+(0-0)^2}\right)^2\)
\(\Rightarrow x^2+y^2=a^2\)
\(\text { So, } x^2+y^2=4 \quad(\because a=2)\)So, \(x^2+y^2=4\)
COMEDK-2011
Hyperbola
120682
If the latus rectum of a hyperbola subtends a right angle at the other focus, then its eccentricity is
120683
Find the coordinates of the foci and the length of the latus rectum of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\).
1 \((0, \pm 2), \frac{32}{3}\)
2 \((0, \pm 5), \frac{32}{3}\)
3 \(( \pm 5,0), \frac{32}{3}\)
4 \((0, \pm 5), \frac{3}{32}\)
Explanation:
C :Given hyperbola,
\(\frac{x^2}{9}-\frac{y^2}{16}=1\)
Here, \(\quad \mathrm{a}=3, \mathrm{~b}=4\) and
\(c=\sqrt{a^2+b^2}=\sqrt{9+16}=5\)Therefore, the coordinates of the foci are \(( \pm 5,0)\)
Also, length of latus rectum \(=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{32}{3}\)
120680
If the eccentricity of a hyperbola is \(5 / 3\), then the eccentricity of its conjugate is
1 \(5 / 3\)
2 \(5 / 4\)
3 5
4 non existent
Explanation:
B Given that, Eccentricity of hyperbola is \(\frac{5}{3}\)
Let \(e_1=\) eccentricity of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
And \(e_2=\) eccentricity of its conjugate \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\)
\(\mathrm{e}_1=\sqrt{\frac{\mathrm{b}^2}{\mathrm{a}^2}+1} \text { and } \mathrm{e}_2=\sqrt{\frac{\mathrm{a}^2}{\mathrm{~b}^2}+1}\)
\(\because \mathrm{e}_1=\frac{5}{3}\)
\(\therefore \frac{25}{9}=\frac{\mathrm{b}^2}{\mathrm{a}^2}+1 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{25}{9}-1=\frac{16}{9}\)
\(\therefore \mathrm{e}_2=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
COMEDK-2011
Hyperbola
120681
The equation of auxiliary circle of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{9}=1\) is
1 \(x^2+y^2=4\)
2 \(x^2+y^2=9\)
3 \(x^2+y^2=13\)
4 \(x^2+y^2=5\)
Explanation:
A Given, hyperbola is \(\frac{\mathrm{x}^2}{4}-\frac{\mathrm{y}^2}{9}=1\)
Auxiliary circle is the circle between the two curves of hyperbola.
\(\therefore(\mathrm{a}, 0)\) and \((-\mathrm{a}, 0)\) touches the circle with centre \((0,0)\)
\(\therefore\) Equation of circle is
\((x-0)^2+(y-0)^2=\left(\sqrt{(a-0)^2+(0-0)^2}\right)^2\)
\(\Rightarrow x^2+y^2=a^2\)
\(\text { So, } x^2+y^2=4 \quad(\because a=2)\)So, \(x^2+y^2=4\)
COMEDK-2011
Hyperbola
120682
If the latus rectum of a hyperbola subtends a right angle at the other focus, then its eccentricity is
120683
Find the coordinates of the foci and the length of the latus rectum of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\).
1 \((0, \pm 2), \frac{32}{3}\)
2 \((0, \pm 5), \frac{32}{3}\)
3 \(( \pm 5,0), \frac{32}{3}\)
4 \((0, \pm 5), \frac{3}{32}\)
Explanation:
C :Given hyperbola,
\(\frac{x^2}{9}-\frac{y^2}{16}=1\)
Here, \(\quad \mathrm{a}=3, \mathrm{~b}=4\) and
\(c=\sqrt{a^2+b^2}=\sqrt{9+16}=5\)Therefore, the coordinates of the foci are \(( \pm 5,0)\)
Also, length of latus rectum \(=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{32}{3}\)
120680
If the eccentricity of a hyperbola is \(5 / 3\), then the eccentricity of its conjugate is
1 \(5 / 3\)
2 \(5 / 4\)
3 5
4 non existent
Explanation:
B Given that, Eccentricity of hyperbola is \(\frac{5}{3}\)
Let \(e_1=\) eccentricity of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
And \(e_2=\) eccentricity of its conjugate \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\)
\(\mathrm{e}_1=\sqrt{\frac{\mathrm{b}^2}{\mathrm{a}^2}+1} \text { and } \mathrm{e}_2=\sqrt{\frac{\mathrm{a}^2}{\mathrm{~b}^2}+1}\)
\(\because \mathrm{e}_1=\frac{5}{3}\)
\(\therefore \frac{25}{9}=\frac{\mathrm{b}^2}{\mathrm{a}^2}+1 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{25}{9}-1=\frac{16}{9}\)
\(\therefore \mathrm{e}_2=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
COMEDK-2011
Hyperbola
120681
The equation of auxiliary circle of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{9}=1\) is
1 \(x^2+y^2=4\)
2 \(x^2+y^2=9\)
3 \(x^2+y^2=13\)
4 \(x^2+y^2=5\)
Explanation:
A Given, hyperbola is \(\frac{\mathrm{x}^2}{4}-\frac{\mathrm{y}^2}{9}=1\)
Auxiliary circle is the circle between the two curves of hyperbola.
\(\therefore(\mathrm{a}, 0)\) and \((-\mathrm{a}, 0)\) touches the circle with centre \((0,0)\)
\(\therefore\) Equation of circle is
\((x-0)^2+(y-0)^2=\left(\sqrt{(a-0)^2+(0-0)^2}\right)^2\)
\(\Rightarrow x^2+y^2=a^2\)
\(\text { So, } x^2+y^2=4 \quad(\because a=2)\)So, \(x^2+y^2=4\)
COMEDK-2011
Hyperbola
120682
If the latus rectum of a hyperbola subtends a right angle at the other focus, then its eccentricity is
120683
Find the coordinates of the foci and the length of the latus rectum of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\).
1 \((0, \pm 2), \frac{32}{3}\)
2 \((0, \pm 5), \frac{32}{3}\)
3 \(( \pm 5,0), \frac{32}{3}\)
4 \((0, \pm 5), \frac{3}{32}\)
Explanation:
C :Given hyperbola,
\(\frac{x^2}{9}-\frac{y^2}{16}=1\)
Here, \(\quad \mathrm{a}=3, \mathrm{~b}=4\) and
\(c=\sqrt{a^2+b^2}=\sqrt{9+16}=5\)Therefore, the coordinates of the foci are \(( \pm 5,0)\)
Also, length of latus rectum \(=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{32}{3}\)
120680
If the eccentricity of a hyperbola is \(5 / 3\), then the eccentricity of its conjugate is
1 \(5 / 3\)
2 \(5 / 4\)
3 5
4 non existent
Explanation:
B Given that, Eccentricity of hyperbola is \(\frac{5}{3}\)
Let \(e_1=\) eccentricity of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
And \(e_2=\) eccentricity of its conjugate \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\)
\(\mathrm{e}_1=\sqrt{\frac{\mathrm{b}^2}{\mathrm{a}^2}+1} \text { and } \mathrm{e}_2=\sqrt{\frac{\mathrm{a}^2}{\mathrm{~b}^2}+1}\)
\(\because \mathrm{e}_1=\frac{5}{3}\)
\(\therefore \frac{25}{9}=\frac{\mathrm{b}^2}{\mathrm{a}^2}+1 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{25}{9}-1=\frac{16}{9}\)
\(\therefore \mathrm{e}_2=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
COMEDK-2011
Hyperbola
120681
The equation of auxiliary circle of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{9}=1\) is
1 \(x^2+y^2=4\)
2 \(x^2+y^2=9\)
3 \(x^2+y^2=13\)
4 \(x^2+y^2=5\)
Explanation:
A Given, hyperbola is \(\frac{\mathrm{x}^2}{4}-\frac{\mathrm{y}^2}{9}=1\)
Auxiliary circle is the circle between the two curves of hyperbola.
\(\therefore(\mathrm{a}, 0)\) and \((-\mathrm{a}, 0)\) touches the circle with centre \((0,0)\)
\(\therefore\) Equation of circle is
\((x-0)^2+(y-0)^2=\left(\sqrt{(a-0)^2+(0-0)^2}\right)^2\)
\(\Rightarrow x^2+y^2=a^2\)
\(\text { So, } x^2+y^2=4 \quad(\because a=2)\)So, \(x^2+y^2=4\)
COMEDK-2011
Hyperbola
120682
If the latus rectum of a hyperbola subtends a right angle at the other focus, then its eccentricity is
120683
Find the coordinates of the foci and the length of the latus rectum of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\).
1 \((0, \pm 2), \frac{32}{3}\)
2 \((0, \pm 5), \frac{32}{3}\)
3 \(( \pm 5,0), \frac{32}{3}\)
4 \((0, \pm 5), \frac{3}{32}\)
Explanation:
C :Given hyperbola,
\(\frac{x^2}{9}-\frac{y^2}{16}=1\)
Here, \(\quad \mathrm{a}=3, \mathrm{~b}=4\) and
\(c=\sqrt{a^2+b^2}=\sqrt{9+16}=5\)Therefore, the coordinates of the foci are \(( \pm 5,0)\)
Also, length of latus rectum \(=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{32}{3}\)
