NEET Test Series from KOTA - 10 Papers In MS WORD
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Ellipse
120564
\(B\) is an extremity of the minor axis of an ellipse whose foci are \(S\) and \(S^{\prime}\). If \(\angle\) SBS' \(^{\prime}\) is a right angle, then the eccentricity of the ellipse is
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{2}{3}\)
4 \(\frac{1}{3}\)
Explanation:
B According to question,
Slope of SB, \(\mathrm{m}_1=\frac{\mathrm{b}-0}{0-\mathrm{ae}}=-\frac{\mathrm{b}}{\mathrm{ae}}\)
and slope of S' B, \(\mathrm{m}_2=\frac{\mathrm{b}-0}{0-(-\mathrm{ae})}=\frac{\mathrm{b}}{\mathrm{ae}}\)
Since, \(\angle\) SBS' is a right angle.
\(\therefore \quad \mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\frac{-\mathrm{b}}{\mathrm{ae}} \times \frac{\mathrm{b}}{\mathrm{ae}}=-1\)
\(b^2=a^2 e^2\)
\(\frac{b^2}{a^2}=e^2\)
\(1-\mathrm{e}^2=\mathrm{e}^2\)
\(2 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{2}\)
\(\mathrm{e}= \pm \frac{1}{\sqrt{2}}\)
\(\mathrm{e}=\frac{1}{\sqrt{2}} \text { or } \mathrm{e}=-\frac{1}{\sqrt{2}}\)
WB JEE-2017
Ellipse
120565
If \(l\) and \(b\) are respectively the length and breadth of the rectangle of greatest area that can be inscribed in the ellipse \(x^2+4 y^2=64\). then \((l, \mathrm{~b})=\)
1 \((16 \sqrt{2}, 4 \sqrt{2})\)
2 \((8 \sqrt{2}, 6 \sqrt{2})\)
3 \((8 \sqrt{2}, 4 \sqrt{2})\)
4 \((6 \sqrt{2}, 4 \sqrt{2})\)
Explanation:
C Given,
\(x^2+4 y^2=64\)
\(\mathrm{x}^2+4 \mathrm{y}^2=64\)
\(\frac{x^2}{64}+\frac{y^2}{16}=1\)
Area of rectangle \(=l \times \mathrm{b}\)
\(A=P Q \times Q R\)
\(A=(2 x) \times(2 y)\)
\(A=4 x y\)
For maximum area \(\frac{\mathrm{dA}}{\mathrm{dx}}=4 \mathrm{y}+4 \mathrm{x} d \mathrm{y}\)
\(\frac{\mathrm{dA}}{\mathrm{dx}}=4 \mathrm{y}+4 \mathrm{x}\left(\frac{-\mathrm{x}}{4 \mathrm{y}}\right)\)
For critical points, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\)
\(4 \mathrm{y}-\frac{\mathrm{x}^2}{\mathrm{y}}=0\)
\(\mathrm{x}= \pm 2 \mathrm{y} \text {, (put in eq }{ }^{\mathrm{n}}(\mathrm{i}) \text { ) }\)
\((2 \mathrm{y})^2+4 \mathrm{y}^2=64\)
\(8 \mathrm{y}^2=64\)
\(\mathrm{y}= \pm 2 \sqrt{2}\)
And \(\quad \mathrm{x}= \pm 4 \sqrt{2}\)
\(\therefore\) Sides will be -
\(\mathrm{PQ}=2 \mathrm{x}=8 \sqrt{2}\)
\(\mathrm{QR}=2 \mathrm{y}=4 \sqrt{2}\)
AP EAMCET-22.04.2019
Ellipse
120566
The eccentricity of the ellipse \(4 x^2+25 y^2=100\) is
1 \(\frac{\sqrt{21}}{5}\)
2 \(\frac{\sqrt{21}}{2}\)
3 \(\frac{\sqrt{21}}{4}\)
4 \(\frac{\sqrt{21}}{25}\)
Explanation:
A Given,
\(4 \mathrm{x}^2+25 \mathrm{y}^2=100\)
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{4}=1\)
So, \(\mathrm{a}^2=25\) and \(\mathrm{b}^2=4\)
Eccentricity \((e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{25}}\)
\(e=\frac{\sqrt{21}}{5}\)
AP EAMCET-22.09.2020
Ellipse
120567
If the angle between the straight lines joining the foci and the ends of the minor axis of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is \(90^{\circ}\), then its eccentricity is
1 \(\frac{1}{2}\)
2 \(\frac{1}{4}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
D Given,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
By Pythagoras theorem -
\(\mathrm{a}^2+\mathrm{a}^2=(2 \mathrm{ae})^2\)
\(2=4 \mathrm{e}^2\)
\(\mathrm{e}^2=\frac{2}{4}\)
\(\mathrm{e}=\frac{1}{\sqrt{2}}\)
120564
\(B\) is an extremity of the minor axis of an ellipse whose foci are \(S\) and \(S^{\prime}\). If \(\angle\) SBS' \(^{\prime}\) is a right angle, then the eccentricity of the ellipse is
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{2}{3}\)
4 \(\frac{1}{3}\)
Explanation:
B According to question,
Slope of SB, \(\mathrm{m}_1=\frac{\mathrm{b}-0}{0-\mathrm{ae}}=-\frac{\mathrm{b}}{\mathrm{ae}}\)
and slope of S' B, \(\mathrm{m}_2=\frac{\mathrm{b}-0}{0-(-\mathrm{ae})}=\frac{\mathrm{b}}{\mathrm{ae}}\)
Since, \(\angle\) SBS' is a right angle.
\(\therefore \quad \mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\frac{-\mathrm{b}}{\mathrm{ae}} \times \frac{\mathrm{b}}{\mathrm{ae}}=-1\)
\(b^2=a^2 e^2\)
\(\frac{b^2}{a^2}=e^2\)
\(1-\mathrm{e}^2=\mathrm{e}^2\)
\(2 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{2}\)
\(\mathrm{e}= \pm \frac{1}{\sqrt{2}}\)
\(\mathrm{e}=\frac{1}{\sqrt{2}} \text { or } \mathrm{e}=-\frac{1}{\sqrt{2}}\)
WB JEE-2017
Ellipse
120565
If \(l\) and \(b\) are respectively the length and breadth of the rectangle of greatest area that can be inscribed in the ellipse \(x^2+4 y^2=64\). then \((l, \mathrm{~b})=\)
1 \((16 \sqrt{2}, 4 \sqrt{2})\)
2 \((8 \sqrt{2}, 6 \sqrt{2})\)
3 \((8 \sqrt{2}, 4 \sqrt{2})\)
4 \((6 \sqrt{2}, 4 \sqrt{2})\)
Explanation:
C Given,
\(x^2+4 y^2=64\)
\(\mathrm{x}^2+4 \mathrm{y}^2=64\)
\(\frac{x^2}{64}+\frac{y^2}{16}=1\)
Area of rectangle \(=l \times \mathrm{b}\)
\(A=P Q \times Q R\)
\(A=(2 x) \times(2 y)\)
\(A=4 x y\)
For maximum area \(\frac{\mathrm{dA}}{\mathrm{dx}}=4 \mathrm{y}+4 \mathrm{x} d \mathrm{y}\)
\(\frac{\mathrm{dA}}{\mathrm{dx}}=4 \mathrm{y}+4 \mathrm{x}\left(\frac{-\mathrm{x}}{4 \mathrm{y}}\right)\)
For critical points, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\)
\(4 \mathrm{y}-\frac{\mathrm{x}^2}{\mathrm{y}}=0\)
\(\mathrm{x}= \pm 2 \mathrm{y} \text {, (put in eq }{ }^{\mathrm{n}}(\mathrm{i}) \text { ) }\)
\((2 \mathrm{y})^2+4 \mathrm{y}^2=64\)
\(8 \mathrm{y}^2=64\)
\(\mathrm{y}= \pm 2 \sqrt{2}\)
And \(\quad \mathrm{x}= \pm 4 \sqrt{2}\)
\(\therefore\) Sides will be -
\(\mathrm{PQ}=2 \mathrm{x}=8 \sqrt{2}\)
\(\mathrm{QR}=2 \mathrm{y}=4 \sqrt{2}\)
AP EAMCET-22.04.2019
Ellipse
120566
The eccentricity of the ellipse \(4 x^2+25 y^2=100\) is
1 \(\frac{\sqrt{21}}{5}\)
2 \(\frac{\sqrt{21}}{2}\)
3 \(\frac{\sqrt{21}}{4}\)
4 \(\frac{\sqrt{21}}{25}\)
Explanation:
A Given,
\(4 \mathrm{x}^2+25 \mathrm{y}^2=100\)
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{4}=1\)
So, \(\mathrm{a}^2=25\) and \(\mathrm{b}^2=4\)
Eccentricity \((e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{25}}\)
\(e=\frac{\sqrt{21}}{5}\)
AP EAMCET-22.09.2020
Ellipse
120567
If the angle between the straight lines joining the foci and the ends of the minor axis of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is \(90^{\circ}\), then its eccentricity is
1 \(\frac{1}{2}\)
2 \(\frac{1}{4}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
D Given,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
By Pythagoras theorem -
\(\mathrm{a}^2+\mathrm{a}^2=(2 \mathrm{ae})^2\)
\(2=4 \mathrm{e}^2\)
\(\mathrm{e}^2=\frac{2}{4}\)
\(\mathrm{e}=\frac{1}{\sqrt{2}}\)
120564
\(B\) is an extremity of the minor axis of an ellipse whose foci are \(S\) and \(S^{\prime}\). If \(\angle\) SBS' \(^{\prime}\) is a right angle, then the eccentricity of the ellipse is
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{2}{3}\)
4 \(\frac{1}{3}\)
Explanation:
B According to question,
Slope of SB, \(\mathrm{m}_1=\frac{\mathrm{b}-0}{0-\mathrm{ae}}=-\frac{\mathrm{b}}{\mathrm{ae}}\)
and slope of S' B, \(\mathrm{m}_2=\frac{\mathrm{b}-0}{0-(-\mathrm{ae})}=\frac{\mathrm{b}}{\mathrm{ae}}\)
Since, \(\angle\) SBS' is a right angle.
\(\therefore \quad \mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\frac{-\mathrm{b}}{\mathrm{ae}} \times \frac{\mathrm{b}}{\mathrm{ae}}=-1\)
\(b^2=a^2 e^2\)
\(\frac{b^2}{a^2}=e^2\)
\(1-\mathrm{e}^2=\mathrm{e}^2\)
\(2 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{2}\)
\(\mathrm{e}= \pm \frac{1}{\sqrt{2}}\)
\(\mathrm{e}=\frac{1}{\sqrt{2}} \text { or } \mathrm{e}=-\frac{1}{\sqrt{2}}\)
WB JEE-2017
Ellipse
120565
If \(l\) and \(b\) are respectively the length and breadth of the rectangle of greatest area that can be inscribed in the ellipse \(x^2+4 y^2=64\). then \((l, \mathrm{~b})=\)
1 \((16 \sqrt{2}, 4 \sqrt{2})\)
2 \((8 \sqrt{2}, 6 \sqrt{2})\)
3 \((8 \sqrt{2}, 4 \sqrt{2})\)
4 \((6 \sqrt{2}, 4 \sqrt{2})\)
Explanation:
C Given,
\(x^2+4 y^2=64\)
\(\mathrm{x}^2+4 \mathrm{y}^2=64\)
\(\frac{x^2}{64}+\frac{y^2}{16}=1\)
Area of rectangle \(=l \times \mathrm{b}\)
\(A=P Q \times Q R\)
\(A=(2 x) \times(2 y)\)
\(A=4 x y\)
For maximum area \(\frac{\mathrm{dA}}{\mathrm{dx}}=4 \mathrm{y}+4 \mathrm{x} d \mathrm{y}\)
\(\frac{\mathrm{dA}}{\mathrm{dx}}=4 \mathrm{y}+4 \mathrm{x}\left(\frac{-\mathrm{x}}{4 \mathrm{y}}\right)\)
For critical points, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\)
\(4 \mathrm{y}-\frac{\mathrm{x}^2}{\mathrm{y}}=0\)
\(\mathrm{x}= \pm 2 \mathrm{y} \text {, (put in eq }{ }^{\mathrm{n}}(\mathrm{i}) \text { ) }\)
\((2 \mathrm{y})^2+4 \mathrm{y}^2=64\)
\(8 \mathrm{y}^2=64\)
\(\mathrm{y}= \pm 2 \sqrt{2}\)
And \(\quad \mathrm{x}= \pm 4 \sqrt{2}\)
\(\therefore\) Sides will be -
\(\mathrm{PQ}=2 \mathrm{x}=8 \sqrt{2}\)
\(\mathrm{QR}=2 \mathrm{y}=4 \sqrt{2}\)
AP EAMCET-22.04.2019
Ellipse
120566
The eccentricity of the ellipse \(4 x^2+25 y^2=100\) is
1 \(\frac{\sqrt{21}}{5}\)
2 \(\frac{\sqrt{21}}{2}\)
3 \(\frac{\sqrt{21}}{4}\)
4 \(\frac{\sqrt{21}}{25}\)
Explanation:
A Given,
\(4 \mathrm{x}^2+25 \mathrm{y}^2=100\)
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{4}=1\)
So, \(\mathrm{a}^2=25\) and \(\mathrm{b}^2=4\)
Eccentricity \((e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{25}}\)
\(e=\frac{\sqrt{21}}{5}\)
AP EAMCET-22.09.2020
Ellipse
120567
If the angle between the straight lines joining the foci and the ends of the minor axis of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is \(90^{\circ}\), then its eccentricity is
1 \(\frac{1}{2}\)
2 \(\frac{1}{4}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
D Given,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
By Pythagoras theorem -
\(\mathrm{a}^2+\mathrm{a}^2=(2 \mathrm{ae})^2\)
\(2=4 \mathrm{e}^2\)
\(\mathrm{e}^2=\frac{2}{4}\)
\(\mathrm{e}=\frac{1}{\sqrt{2}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ellipse
120564
\(B\) is an extremity of the minor axis of an ellipse whose foci are \(S\) and \(S^{\prime}\). If \(\angle\) SBS' \(^{\prime}\) is a right angle, then the eccentricity of the ellipse is
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{2}{3}\)
4 \(\frac{1}{3}\)
Explanation:
B According to question,
Slope of SB, \(\mathrm{m}_1=\frac{\mathrm{b}-0}{0-\mathrm{ae}}=-\frac{\mathrm{b}}{\mathrm{ae}}\)
and slope of S' B, \(\mathrm{m}_2=\frac{\mathrm{b}-0}{0-(-\mathrm{ae})}=\frac{\mathrm{b}}{\mathrm{ae}}\)
Since, \(\angle\) SBS' is a right angle.
\(\therefore \quad \mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\frac{-\mathrm{b}}{\mathrm{ae}} \times \frac{\mathrm{b}}{\mathrm{ae}}=-1\)
\(b^2=a^2 e^2\)
\(\frac{b^2}{a^2}=e^2\)
\(1-\mathrm{e}^2=\mathrm{e}^2\)
\(2 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{2}\)
\(\mathrm{e}= \pm \frac{1}{\sqrt{2}}\)
\(\mathrm{e}=\frac{1}{\sqrt{2}} \text { or } \mathrm{e}=-\frac{1}{\sqrt{2}}\)
WB JEE-2017
Ellipse
120565
If \(l\) and \(b\) are respectively the length and breadth of the rectangle of greatest area that can be inscribed in the ellipse \(x^2+4 y^2=64\). then \((l, \mathrm{~b})=\)
1 \((16 \sqrt{2}, 4 \sqrt{2})\)
2 \((8 \sqrt{2}, 6 \sqrt{2})\)
3 \((8 \sqrt{2}, 4 \sqrt{2})\)
4 \((6 \sqrt{2}, 4 \sqrt{2})\)
Explanation:
C Given,
\(x^2+4 y^2=64\)
\(\mathrm{x}^2+4 \mathrm{y}^2=64\)
\(\frac{x^2}{64}+\frac{y^2}{16}=1\)
Area of rectangle \(=l \times \mathrm{b}\)
\(A=P Q \times Q R\)
\(A=(2 x) \times(2 y)\)
\(A=4 x y\)
For maximum area \(\frac{\mathrm{dA}}{\mathrm{dx}}=4 \mathrm{y}+4 \mathrm{x} d \mathrm{y}\)
\(\frac{\mathrm{dA}}{\mathrm{dx}}=4 \mathrm{y}+4 \mathrm{x}\left(\frac{-\mathrm{x}}{4 \mathrm{y}}\right)\)
For critical points, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\)
\(4 \mathrm{y}-\frac{\mathrm{x}^2}{\mathrm{y}}=0\)
\(\mathrm{x}= \pm 2 \mathrm{y} \text {, (put in eq }{ }^{\mathrm{n}}(\mathrm{i}) \text { ) }\)
\((2 \mathrm{y})^2+4 \mathrm{y}^2=64\)
\(8 \mathrm{y}^2=64\)
\(\mathrm{y}= \pm 2 \sqrt{2}\)
And \(\quad \mathrm{x}= \pm 4 \sqrt{2}\)
\(\therefore\) Sides will be -
\(\mathrm{PQ}=2 \mathrm{x}=8 \sqrt{2}\)
\(\mathrm{QR}=2 \mathrm{y}=4 \sqrt{2}\)
AP EAMCET-22.04.2019
Ellipse
120566
The eccentricity of the ellipse \(4 x^2+25 y^2=100\) is
1 \(\frac{\sqrt{21}}{5}\)
2 \(\frac{\sqrt{21}}{2}\)
3 \(\frac{\sqrt{21}}{4}\)
4 \(\frac{\sqrt{21}}{25}\)
Explanation:
A Given,
\(4 \mathrm{x}^2+25 \mathrm{y}^2=100\)
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{4}=1\)
So, \(\mathrm{a}^2=25\) and \(\mathrm{b}^2=4\)
Eccentricity \((e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{25}}\)
\(e=\frac{\sqrt{21}}{5}\)
AP EAMCET-22.09.2020
Ellipse
120567
If the angle between the straight lines joining the foci and the ends of the minor axis of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is \(90^{\circ}\), then its eccentricity is
1 \(\frac{1}{2}\)
2 \(\frac{1}{4}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
D Given,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
By Pythagoras theorem -
\(\mathrm{a}^2+\mathrm{a}^2=(2 \mathrm{ae})^2\)
\(2=4 \mathrm{e}^2\)
\(\mathrm{e}^2=\frac{2}{4}\)
\(\mathrm{e}=\frac{1}{\sqrt{2}}\)
