120732
The length of conjugate axis of a hyperbola is greater than the length of transverse axis.
Then the eccentricity e is
1 \(=\sqrt{2}\)
2 \(>\sqrt{2}\)
3 \(\lt \sqrt{2}\)
4 \(\lt \frac{1}{\sqrt{2}}\)
Explanation:
B Let the length of conjugate axis \(=\mathrm{b}\) and the length of transverse axis \(=\mathrm{a}\)
Given that, \(\mathrm{b}>\mathrm{a}\)
Now, \(\mathrm{b}^2>\mathrm{a}^2\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}>1\)
On adding 1 on both the side we get -
\(1+\frac{\mathrm{b}^2}{\mathrm{a}^2}>2\)
\(\mathrm{e}^2>2\)
\(\mathrm{e}>\sqrt{2}\)
\(\left[\because \sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\mathrm{e}\right]\)
WB JEE-2019
Hyperbola
120733
The equation of the directrices of the hyperbola \(3 x^2-3 y^2-18 x+12 y+2=0\) is
1 \(x=3 \pm \sqrt{\frac{13}{6}}\)
2 \(x=3 \pm \sqrt{\frac{6}{13}}\)
3 \(x=6 \pm \sqrt{\frac{13}{3}}\)
4 \(x=6 \pm \sqrt{\frac{3}{13}}\)
Explanation:
A Given equation is
\(3 x^2-3 y^2-18 x+12 y+2=0 .\)
It can be written as
\(\frac{(x-3)^2}{\left(\sqrt{\frac{13}{3}}\right)^2}-\frac{(y-3)^2}{\left(\sqrt{\frac{13}{3}}\right)^2}=1\)
here, \(a=b=\sqrt{\frac{13}{3}}\)
\(\therefore \quad \mathrm{e} =\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1+1}\)
\(=\sqrt{2}\)
\(\therefore\) Equation of the directrix are
\(x=3 \pm \sqrt{\frac{13}{6}}\)
WB JEE-2019
Hyperbola
120734
If the latus rectum of a hyperbola through one focus subtends an angle \(60^{\circ}\) at the other focus, then its eccentricity is.....
1 \(\sqrt{2}\)
2 \(\sqrt{6}\)
3 \(\sqrt{3}\)
4 \(\sqrt{5}\)
Explanation:
C Given,
Latus rectum subtends angle \(60^{\circ}\) at focus.
Let foci of hyperbola \(\mathrm{S}( \pm \mathrm{ae}, 0)\)
and points of latus rectum (ae, \(\pm b^2 / a\) )
Angle between L. \(\mathrm{R}\) an \(\mathrm{x}\)-axis is \(30^{\circ}\)
Slope, \(\mathrm{m}_1=\frac{\mathrm{b}^2}{2 \mathrm{a}^2 \mathrm{e}}=\tan 30^{\circ}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{2 \mathrm{e}}{\sqrt{3}}=\mathrm{e}^2-1\)
\(\sqrt{3} \mathrm{e}^2-\sqrt{3}-2 \mathrm{e}=0\)
\(\sqrt{3} \mathrm{e}^2-2 \mathrm{e}-\sqrt{3}=0\)
\((\mathrm{e}-\sqrt{3})(\sqrt{2} \mathrm{e}-1)=0\)
\(\mathrm{e}=\sqrt{3} \text { or } \frac{-1}{\sqrt{2}}\)
Hence,
\(\mathrm{e}=\sqrt{3}\)
AP EAMCET-17.09.2020
Hyperbola
120735
The vertices of the hyperbola \(7 x^2-49 y^2=343\) having eccentricity ' \(4 / 3\) ' is
1 \((0,0)\)
2 \(( \pm 3,0)\)
3 \((0, \pm 5)\)
4 \(( \pm 7,0)\)
Explanation:
D Given,
\(7 x^2-49 y^2=343\)
\(\frac{x^2}{49}-\frac{y^2}{7}=1\)
Here, \(\quad \mathrm{a}^2=49\) and \(\mathrm{b}^2=7\)
Then, its vertices are \(( \pm \mathrm{a}, 0)=( \pm 7,0)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Hyperbola
120732
The length of conjugate axis of a hyperbola is greater than the length of transverse axis.
Then the eccentricity e is
1 \(=\sqrt{2}\)
2 \(>\sqrt{2}\)
3 \(\lt \sqrt{2}\)
4 \(\lt \frac{1}{\sqrt{2}}\)
Explanation:
B Let the length of conjugate axis \(=\mathrm{b}\) and the length of transverse axis \(=\mathrm{a}\)
Given that, \(\mathrm{b}>\mathrm{a}\)
Now, \(\mathrm{b}^2>\mathrm{a}^2\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}>1\)
On adding 1 on both the side we get -
\(1+\frac{\mathrm{b}^2}{\mathrm{a}^2}>2\)
\(\mathrm{e}^2>2\)
\(\mathrm{e}>\sqrt{2}\)
\(\left[\because \sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\mathrm{e}\right]\)
WB JEE-2019
Hyperbola
120733
The equation of the directrices of the hyperbola \(3 x^2-3 y^2-18 x+12 y+2=0\) is
1 \(x=3 \pm \sqrt{\frac{13}{6}}\)
2 \(x=3 \pm \sqrt{\frac{6}{13}}\)
3 \(x=6 \pm \sqrt{\frac{13}{3}}\)
4 \(x=6 \pm \sqrt{\frac{3}{13}}\)
Explanation:
A Given equation is
\(3 x^2-3 y^2-18 x+12 y+2=0 .\)
It can be written as
\(\frac{(x-3)^2}{\left(\sqrt{\frac{13}{3}}\right)^2}-\frac{(y-3)^2}{\left(\sqrt{\frac{13}{3}}\right)^2}=1\)
here, \(a=b=\sqrt{\frac{13}{3}}\)
\(\therefore \quad \mathrm{e} =\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1+1}\)
\(=\sqrt{2}\)
\(\therefore\) Equation of the directrix are
\(x=3 \pm \sqrt{\frac{13}{6}}\)
WB JEE-2019
Hyperbola
120734
If the latus rectum of a hyperbola through one focus subtends an angle \(60^{\circ}\) at the other focus, then its eccentricity is.....
1 \(\sqrt{2}\)
2 \(\sqrt{6}\)
3 \(\sqrt{3}\)
4 \(\sqrt{5}\)
Explanation:
C Given,
Latus rectum subtends angle \(60^{\circ}\) at focus.
Let foci of hyperbola \(\mathrm{S}( \pm \mathrm{ae}, 0)\)
and points of latus rectum (ae, \(\pm b^2 / a\) )
Angle between L. \(\mathrm{R}\) an \(\mathrm{x}\)-axis is \(30^{\circ}\)
Slope, \(\mathrm{m}_1=\frac{\mathrm{b}^2}{2 \mathrm{a}^2 \mathrm{e}}=\tan 30^{\circ}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{2 \mathrm{e}}{\sqrt{3}}=\mathrm{e}^2-1\)
\(\sqrt{3} \mathrm{e}^2-\sqrt{3}-2 \mathrm{e}=0\)
\(\sqrt{3} \mathrm{e}^2-2 \mathrm{e}-\sqrt{3}=0\)
\((\mathrm{e}-\sqrt{3})(\sqrt{2} \mathrm{e}-1)=0\)
\(\mathrm{e}=\sqrt{3} \text { or } \frac{-1}{\sqrt{2}}\)
Hence,
\(\mathrm{e}=\sqrt{3}\)
AP EAMCET-17.09.2020
Hyperbola
120735
The vertices of the hyperbola \(7 x^2-49 y^2=343\) having eccentricity ' \(4 / 3\) ' is
1 \((0,0)\)
2 \(( \pm 3,0)\)
3 \((0, \pm 5)\)
4 \(( \pm 7,0)\)
Explanation:
D Given,
\(7 x^2-49 y^2=343\)
\(\frac{x^2}{49}-\frac{y^2}{7}=1\)
Here, \(\quad \mathrm{a}^2=49\) and \(\mathrm{b}^2=7\)
Then, its vertices are \(( \pm \mathrm{a}, 0)=( \pm 7,0)\)
120732
The length of conjugate axis of a hyperbola is greater than the length of transverse axis.
Then the eccentricity e is
1 \(=\sqrt{2}\)
2 \(>\sqrt{2}\)
3 \(\lt \sqrt{2}\)
4 \(\lt \frac{1}{\sqrt{2}}\)
Explanation:
B Let the length of conjugate axis \(=\mathrm{b}\) and the length of transverse axis \(=\mathrm{a}\)
Given that, \(\mathrm{b}>\mathrm{a}\)
Now, \(\mathrm{b}^2>\mathrm{a}^2\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}>1\)
On adding 1 on both the side we get -
\(1+\frac{\mathrm{b}^2}{\mathrm{a}^2}>2\)
\(\mathrm{e}^2>2\)
\(\mathrm{e}>\sqrt{2}\)
\(\left[\because \sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\mathrm{e}\right]\)
WB JEE-2019
Hyperbola
120733
The equation of the directrices of the hyperbola \(3 x^2-3 y^2-18 x+12 y+2=0\) is
1 \(x=3 \pm \sqrt{\frac{13}{6}}\)
2 \(x=3 \pm \sqrt{\frac{6}{13}}\)
3 \(x=6 \pm \sqrt{\frac{13}{3}}\)
4 \(x=6 \pm \sqrt{\frac{3}{13}}\)
Explanation:
A Given equation is
\(3 x^2-3 y^2-18 x+12 y+2=0 .\)
It can be written as
\(\frac{(x-3)^2}{\left(\sqrt{\frac{13}{3}}\right)^2}-\frac{(y-3)^2}{\left(\sqrt{\frac{13}{3}}\right)^2}=1\)
here, \(a=b=\sqrt{\frac{13}{3}}\)
\(\therefore \quad \mathrm{e} =\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1+1}\)
\(=\sqrt{2}\)
\(\therefore\) Equation of the directrix are
\(x=3 \pm \sqrt{\frac{13}{6}}\)
WB JEE-2019
Hyperbola
120734
If the latus rectum of a hyperbola through one focus subtends an angle \(60^{\circ}\) at the other focus, then its eccentricity is.....
1 \(\sqrt{2}\)
2 \(\sqrt{6}\)
3 \(\sqrt{3}\)
4 \(\sqrt{5}\)
Explanation:
C Given,
Latus rectum subtends angle \(60^{\circ}\) at focus.
Let foci of hyperbola \(\mathrm{S}( \pm \mathrm{ae}, 0)\)
and points of latus rectum (ae, \(\pm b^2 / a\) )
Angle between L. \(\mathrm{R}\) an \(\mathrm{x}\)-axis is \(30^{\circ}\)
Slope, \(\mathrm{m}_1=\frac{\mathrm{b}^2}{2 \mathrm{a}^2 \mathrm{e}}=\tan 30^{\circ}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{2 \mathrm{e}}{\sqrt{3}}=\mathrm{e}^2-1\)
\(\sqrt{3} \mathrm{e}^2-\sqrt{3}-2 \mathrm{e}=0\)
\(\sqrt{3} \mathrm{e}^2-2 \mathrm{e}-\sqrt{3}=0\)
\((\mathrm{e}-\sqrt{3})(\sqrt{2} \mathrm{e}-1)=0\)
\(\mathrm{e}=\sqrt{3} \text { or } \frac{-1}{\sqrt{2}}\)
Hence,
\(\mathrm{e}=\sqrt{3}\)
AP EAMCET-17.09.2020
Hyperbola
120735
The vertices of the hyperbola \(7 x^2-49 y^2=343\) having eccentricity ' \(4 / 3\) ' is
1 \((0,0)\)
2 \(( \pm 3,0)\)
3 \((0, \pm 5)\)
4 \(( \pm 7,0)\)
Explanation:
D Given,
\(7 x^2-49 y^2=343\)
\(\frac{x^2}{49}-\frac{y^2}{7}=1\)
Here, \(\quad \mathrm{a}^2=49\) and \(\mathrm{b}^2=7\)
Then, its vertices are \(( \pm \mathrm{a}, 0)=( \pm 7,0)\)
120732
The length of conjugate axis of a hyperbola is greater than the length of transverse axis.
Then the eccentricity e is
1 \(=\sqrt{2}\)
2 \(>\sqrt{2}\)
3 \(\lt \sqrt{2}\)
4 \(\lt \frac{1}{\sqrt{2}}\)
Explanation:
B Let the length of conjugate axis \(=\mathrm{b}\) and the length of transverse axis \(=\mathrm{a}\)
Given that, \(\mathrm{b}>\mathrm{a}\)
Now, \(\mathrm{b}^2>\mathrm{a}^2\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}>1\)
On adding 1 on both the side we get -
\(1+\frac{\mathrm{b}^2}{\mathrm{a}^2}>2\)
\(\mathrm{e}^2>2\)
\(\mathrm{e}>\sqrt{2}\)
\(\left[\because \sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\mathrm{e}\right]\)
WB JEE-2019
Hyperbola
120733
The equation of the directrices of the hyperbola \(3 x^2-3 y^2-18 x+12 y+2=0\) is
1 \(x=3 \pm \sqrt{\frac{13}{6}}\)
2 \(x=3 \pm \sqrt{\frac{6}{13}}\)
3 \(x=6 \pm \sqrt{\frac{13}{3}}\)
4 \(x=6 \pm \sqrt{\frac{3}{13}}\)
Explanation:
A Given equation is
\(3 x^2-3 y^2-18 x+12 y+2=0 .\)
It can be written as
\(\frac{(x-3)^2}{\left(\sqrt{\frac{13}{3}}\right)^2}-\frac{(y-3)^2}{\left(\sqrt{\frac{13}{3}}\right)^2}=1\)
here, \(a=b=\sqrt{\frac{13}{3}}\)
\(\therefore \quad \mathrm{e} =\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(=\sqrt{1+1}\)
\(=\sqrt{2}\)
\(\therefore\) Equation of the directrix are
\(x=3 \pm \sqrt{\frac{13}{6}}\)
WB JEE-2019
Hyperbola
120734
If the latus rectum of a hyperbola through one focus subtends an angle \(60^{\circ}\) at the other focus, then its eccentricity is.....
1 \(\sqrt{2}\)
2 \(\sqrt{6}\)
3 \(\sqrt{3}\)
4 \(\sqrt{5}\)
Explanation:
C Given,
Latus rectum subtends angle \(60^{\circ}\) at focus.
Let foci of hyperbola \(\mathrm{S}( \pm \mathrm{ae}, 0)\)
and points of latus rectum (ae, \(\pm b^2 / a\) )
Angle between L. \(\mathrm{R}\) an \(\mathrm{x}\)-axis is \(30^{\circ}\)
Slope, \(\mathrm{m}_1=\frac{\mathrm{b}^2}{2 \mathrm{a}^2 \mathrm{e}}=\tan 30^{\circ}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{2 \mathrm{e}}{\sqrt{3}}=\mathrm{e}^2-1\)
\(\sqrt{3} \mathrm{e}^2-\sqrt{3}-2 \mathrm{e}=0\)
\(\sqrt{3} \mathrm{e}^2-2 \mathrm{e}-\sqrt{3}=0\)
\((\mathrm{e}-\sqrt{3})(\sqrt{2} \mathrm{e}-1)=0\)
\(\mathrm{e}=\sqrt{3} \text { or } \frac{-1}{\sqrt{2}}\)
Hence,
\(\mathrm{e}=\sqrt{3}\)
AP EAMCET-17.09.2020
Hyperbola
120735
The vertices of the hyperbola \(7 x^2-49 y^2=343\) having eccentricity ' \(4 / 3\) ' is
1 \((0,0)\)
2 \(( \pm 3,0)\)
3 \((0, \pm 5)\)
4 \(( \pm 7,0)\)
Explanation:
D Given,
\(7 x^2-49 y^2=343\)
\(\frac{x^2}{49}-\frac{y^2}{7}=1\)
Here, \(\quad \mathrm{a}^2=49\) and \(\mathrm{b}^2=7\)
Then, its vertices are \(( \pm \mathrm{a}, 0)=( \pm 7,0)\)