NEET Test Series from KOTA - 10 Papers In MS WORD
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Ellipse
120543
The major and minor axes of an ellipse are along the \(\mathrm{X}\)-axis and \(\mathrm{Y}\)-axis respectively. If its latusrectum is of length 4 and the distance between the foci is \(4 \sqrt{2}\), then the equation of than ellipse is
120544
The distance between the focii of the ellipse \(x=\) \(3 \cos \theta, \mathrm{y}=4 \sin \theta\) is
1 \(2 \sqrt{7}\)
2 \(7 \sqrt{2}\)
3 \(\sqrt{7}\)
4 \(3 \sqrt{7}\)
Explanation:
A Given ellipse : \(\mathrm{x}=3 \cos \theta, \mathrm{y}=4 \sin \theta\)
Here, \(a=3\) and \(b=4\)
For horizontal ellipse \(\mathrm{b}\lt \mathrm{a}\)
And for vertical ellipse, \(b>a\)
So, we can interchange it
Like, \(a=4\) and \(b=3\)
Also, \(\quad b^2=a^2\left(1-e^2\right)\)
\(9=16\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}=\frac{\sqrt{7}}{4}\)
Distance between two foci \(=2 \mathrm{ae}\)
\(\mathrm{d}=2 \times 4 \times \frac{\sqrt{7}}{4}\)
\(\mathrm{~d}=2 \sqrt{7}\)
AP EAMCET-2016
Ellipse
120545
Foci of the ellipse
\(2 x^2+3 y^2-4 x-12 y+13=0\) are
1 \(\left(1+\frac{1}{\sqrt{6}}, 2\right)\) and \(\left(1-\frac{1}{\sqrt{6}}, 2\right)\)
2 \(\left(\frac{1}{\sqrt{6}}+1,2\right)\) and \(\left(\frac{1}{\sqrt{6}}-1,2\right)\)
3 \(\left(2,1+\frac{1}{\sqrt{6}}+1\right)\) and \(\left(2,1-\frac{1}{\sqrt{6}}\right)\)
4 \(\left(2, \frac{1}{\sqrt{6}}+1\right)\) and \(\left(2, \frac{1}{\sqrt{6}}-1\right)\)
Explanation:
A Given ellipse,
\(2 \mathrm{x}^2+3 \mathrm{y}^2-4 \mathrm{x}-12 \mathrm{y}+13=0\)
\(2(\mathrm{x}-1)^2+3(\mathrm{y}-2)^2=1\)
\(\frac{(x-1)^2}{\left(\frac{1}{\sqrt{2}}\right)^2}+\frac{(y-2)^2}{\left(\frac{1}{\sqrt{3}}\right)^2}=1\)
Centre of the ellipse \(=(\mathrm{h}, \mathrm{k}) \equiv(1,2)\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{1 / 3}{1 / 2}}=\frac{1}{\sqrt{3}}\)
\(\mathrm{a}=\frac{1}{\sqrt{2}}, \mathrm{~b}=\frac{1}{\sqrt{3}} \Rightarrow \mathrm{ae}=\frac{1}{\sqrt{6}}\)Foci \(\quad(\mathrm{h} \pm \mathrm{ae}, \mathrm{k}) \equiv\left(1 \pm \frac{1}{\sqrt{6}}, 2\right)\)
AP EAMCET-17.09.2020
Ellipse
120546
If the line joining the points \(A(\alpha)\) and \(B(\beta)\) on the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) is focal chord, then one possible value of \(\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2}\) is
1 -3
2 3
3 -9
4 9
Explanation:
C Given, ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\)
Here, \(\mathrm{a}=5\) and \(\mathrm{b}=3\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}\)
Equation of chord joining \(A(\alpha)\) and \(B(\beta)\) is
\(\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)\)
It passes through focus \((\mathrm{ae}, 0)\)
\(\frac{\mathrm{ae}}{\mathrm{a}} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{0}{\mathrm{~b}} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)\)
\(\frac{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}{\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{1}{\mathrm{e}}\)
\(\frac{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)+\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)-\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{1+\mathrm{e}}{1-\mathrm{e}}\)
\(\frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{-2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=\frac{1+\mathrm{e}}{1-\mathrm{e}}\)
\(\cot \left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right)=\frac{e+1}{e-1}=\frac{\frac{4}{5}+1}{\frac{4}{5}-1}=-9\)
\(\cot \left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right)=-9\)
120543
The major and minor axes of an ellipse are along the \(\mathrm{X}\)-axis and \(\mathrm{Y}\)-axis respectively. If its latusrectum is of length 4 and the distance between the foci is \(4 \sqrt{2}\), then the equation of than ellipse is
120544
The distance between the focii of the ellipse \(x=\) \(3 \cos \theta, \mathrm{y}=4 \sin \theta\) is
1 \(2 \sqrt{7}\)
2 \(7 \sqrt{2}\)
3 \(\sqrt{7}\)
4 \(3 \sqrt{7}\)
Explanation:
A Given ellipse : \(\mathrm{x}=3 \cos \theta, \mathrm{y}=4 \sin \theta\)
Here, \(a=3\) and \(b=4\)
For horizontal ellipse \(\mathrm{b}\lt \mathrm{a}\)
And for vertical ellipse, \(b>a\)
So, we can interchange it
Like, \(a=4\) and \(b=3\)
Also, \(\quad b^2=a^2\left(1-e^2\right)\)
\(9=16\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}=\frac{\sqrt{7}}{4}\)
Distance between two foci \(=2 \mathrm{ae}\)
\(\mathrm{d}=2 \times 4 \times \frac{\sqrt{7}}{4}\)
\(\mathrm{~d}=2 \sqrt{7}\)
AP EAMCET-2016
Ellipse
120545
Foci of the ellipse
\(2 x^2+3 y^2-4 x-12 y+13=0\) are
1 \(\left(1+\frac{1}{\sqrt{6}}, 2\right)\) and \(\left(1-\frac{1}{\sqrt{6}}, 2\right)\)
2 \(\left(\frac{1}{\sqrt{6}}+1,2\right)\) and \(\left(\frac{1}{\sqrt{6}}-1,2\right)\)
3 \(\left(2,1+\frac{1}{\sqrt{6}}+1\right)\) and \(\left(2,1-\frac{1}{\sqrt{6}}\right)\)
4 \(\left(2, \frac{1}{\sqrt{6}}+1\right)\) and \(\left(2, \frac{1}{\sqrt{6}}-1\right)\)
Explanation:
A Given ellipse,
\(2 \mathrm{x}^2+3 \mathrm{y}^2-4 \mathrm{x}-12 \mathrm{y}+13=0\)
\(2(\mathrm{x}-1)^2+3(\mathrm{y}-2)^2=1\)
\(\frac{(x-1)^2}{\left(\frac{1}{\sqrt{2}}\right)^2}+\frac{(y-2)^2}{\left(\frac{1}{\sqrt{3}}\right)^2}=1\)
Centre of the ellipse \(=(\mathrm{h}, \mathrm{k}) \equiv(1,2)\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{1 / 3}{1 / 2}}=\frac{1}{\sqrt{3}}\)
\(\mathrm{a}=\frac{1}{\sqrt{2}}, \mathrm{~b}=\frac{1}{\sqrt{3}} \Rightarrow \mathrm{ae}=\frac{1}{\sqrt{6}}\)Foci \(\quad(\mathrm{h} \pm \mathrm{ae}, \mathrm{k}) \equiv\left(1 \pm \frac{1}{\sqrt{6}}, 2\right)\)
AP EAMCET-17.09.2020
Ellipse
120546
If the line joining the points \(A(\alpha)\) and \(B(\beta)\) on the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) is focal chord, then one possible value of \(\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2}\) is
1 -3
2 3
3 -9
4 9
Explanation:
C Given, ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\)
Here, \(\mathrm{a}=5\) and \(\mathrm{b}=3\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}\)
Equation of chord joining \(A(\alpha)\) and \(B(\beta)\) is
\(\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)\)
It passes through focus \((\mathrm{ae}, 0)\)
\(\frac{\mathrm{ae}}{\mathrm{a}} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{0}{\mathrm{~b}} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)\)
\(\frac{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}{\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{1}{\mathrm{e}}\)
\(\frac{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)+\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)-\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{1+\mathrm{e}}{1-\mathrm{e}}\)
\(\frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{-2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=\frac{1+\mathrm{e}}{1-\mathrm{e}}\)
\(\cot \left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right)=\frac{e+1}{e-1}=\frac{\frac{4}{5}+1}{\frac{4}{5}-1}=-9\)
\(\cot \left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right)=-9\)
120543
The major and minor axes of an ellipse are along the \(\mathrm{X}\)-axis and \(\mathrm{Y}\)-axis respectively. If its latusrectum is of length 4 and the distance between the foci is \(4 \sqrt{2}\), then the equation of than ellipse is
120544
The distance between the focii of the ellipse \(x=\) \(3 \cos \theta, \mathrm{y}=4 \sin \theta\) is
1 \(2 \sqrt{7}\)
2 \(7 \sqrt{2}\)
3 \(\sqrt{7}\)
4 \(3 \sqrt{7}\)
Explanation:
A Given ellipse : \(\mathrm{x}=3 \cos \theta, \mathrm{y}=4 \sin \theta\)
Here, \(a=3\) and \(b=4\)
For horizontal ellipse \(\mathrm{b}\lt \mathrm{a}\)
And for vertical ellipse, \(b>a\)
So, we can interchange it
Like, \(a=4\) and \(b=3\)
Also, \(\quad b^2=a^2\left(1-e^2\right)\)
\(9=16\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}=\frac{\sqrt{7}}{4}\)
Distance between two foci \(=2 \mathrm{ae}\)
\(\mathrm{d}=2 \times 4 \times \frac{\sqrt{7}}{4}\)
\(\mathrm{~d}=2 \sqrt{7}\)
AP EAMCET-2016
Ellipse
120545
Foci of the ellipse
\(2 x^2+3 y^2-4 x-12 y+13=0\) are
1 \(\left(1+\frac{1}{\sqrt{6}}, 2\right)\) and \(\left(1-\frac{1}{\sqrt{6}}, 2\right)\)
2 \(\left(\frac{1}{\sqrt{6}}+1,2\right)\) and \(\left(\frac{1}{\sqrt{6}}-1,2\right)\)
3 \(\left(2,1+\frac{1}{\sqrt{6}}+1\right)\) and \(\left(2,1-\frac{1}{\sqrt{6}}\right)\)
4 \(\left(2, \frac{1}{\sqrt{6}}+1\right)\) and \(\left(2, \frac{1}{\sqrt{6}}-1\right)\)
Explanation:
A Given ellipse,
\(2 \mathrm{x}^2+3 \mathrm{y}^2-4 \mathrm{x}-12 \mathrm{y}+13=0\)
\(2(\mathrm{x}-1)^2+3(\mathrm{y}-2)^2=1\)
\(\frac{(x-1)^2}{\left(\frac{1}{\sqrt{2}}\right)^2}+\frac{(y-2)^2}{\left(\frac{1}{\sqrt{3}}\right)^2}=1\)
Centre of the ellipse \(=(\mathrm{h}, \mathrm{k}) \equiv(1,2)\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{1 / 3}{1 / 2}}=\frac{1}{\sqrt{3}}\)
\(\mathrm{a}=\frac{1}{\sqrt{2}}, \mathrm{~b}=\frac{1}{\sqrt{3}} \Rightarrow \mathrm{ae}=\frac{1}{\sqrt{6}}\)Foci \(\quad(\mathrm{h} \pm \mathrm{ae}, \mathrm{k}) \equiv\left(1 \pm \frac{1}{\sqrt{6}}, 2\right)\)
AP EAMCET-17.09.2020
Ellipse
120546
If the line joining the points \(A(\alpha)\) and \(B(\beta)\) on the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) is focal chord, then one possible value of \(\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2}\) is
1 -3
2 3
3 -9
4 9
Explanation:
C Given, ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\)
Here, \(\mathrm{a}=5\) and \(\mathrm{b}=3\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}\)
Equation of chord joining \(A(\alpha)\) and \(B(\beta)\) is
\(\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)\)
It passes through focus \((\mathrm{ae}, 0)\)
\(\frac{\mathrm{ae}}{\mathrm{a}} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{0}{\mathrm{~b}} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)\)
\(\frac{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}{\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{1}{\mathrm{e}}\)
\(\frac{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)+\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)-\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{1+\mathrm{e}}{1-\mathrm{e}}\)
\(\frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{-2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=\frac{1+\mathrm{e}}{1-\mathrm{e}}\)
\(\cot \left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right)=\frac{e+1}{e-1}=\frac{\frac{4}{5}+1}{\frac{4}{5}-1}=-9\)
\(\cot \left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right)=-9\)
120543
The major and minor axes of an ellipse are along the \(\mathrm{X}\)-axis and \(\mathrm{Y}\)-axis respectively. If its latusrectum is of length 4 and the distance between the foci is \(4 \sqrt{2}\), then the equation of than ellipse is
120544
The distance between the focii of the ellipse \(x=\) \(3 \cos \theta, \mathrm{y}=4 \sin \theta\) is
1 \(2 \sqrt{7}\)
2 \(7 \sqrt{2}\)
3 \(\sqrt{7}\)
4 \(3 \sqrt{7}\)
Explanation:
A Given ellipse : \(\mathrm{x}=3 \cos \theta, \mathrm{y}=4 \sin \theta\)
Here, \(a=3\) and \(b=4\)
For horizontal ellipse \(\mathrm{b}\lt \mathrm{a}\)
And for vertical ellipse, \(b>a\)
So, we can interchange it
Like, \(a=4\) and \(b=3\)
Also, \(\quad b^2=a^2\left(1-e^2\right)\)
\(9=16\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}=\frac{\sqrt{7}}{4}\)
Distance between two foci \(=2 \mathrm{ae}\)
\(\mathrm{d}=2 \times 4 \times \frac{\sqrt{7}}{4}\)
\(\mathrm{~d}=2 \sqrt{7}\)
AP EAMCET-2016
Ellipse
120545
Foci of the ellipse
\(2 x^2+3 y^2-4 x-12 y+13=0\) are
1 \(\left(1+\frac{1}{\sqrt{6}}, 2\right)\) and \(\left(1-\frac{1}{\sqrt{6}}, 2\right)\)
2 \(\left(\frac{1}{\sqrt{6}}+1,2\right)\) and \(\left(\frac{1}{\sqrt{6}}-1,2\right)\)
3 \(\left(2,1+\frac{1}{\sqrt{6}}+1\right)\) and \(\left(2,1-\frac{1}{\sqrt{6}}\right)\)
4 \(\left(2, \frac{1}{\sqrt{6}}+1\right)\) and \(\left(2, \frac{1}{\sqrt{6}}-1\right)\)
Explanation:
A Given ellipse,
\(2 \mathrm{x}^2+3 \mathrm{y}^2-4 \mathrm{x}-12 \mathrm{y}+13=0\)
\(2(\mathrm{x}-1)^2+3(\mathrm{y}-2)^2=1\)
\(\frac{(x-1)^2}{\left(\frac{1}{\sqrt{2}}\right)^2}+\frac{(y-2)^2}{\left(\frac{1}{\sqrt{3}}\right)^2}=1\)
Centre of the ellipse \(=(\mathrm{h}, \mathrm{k}) \equiv(1,2)\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{1 / 3}{1 / 2}}=\frac{1}{\sqrt{3}}\)
\(\mathrm{a}=\frac{1}{\sqrt{2}}, \mathrm{~b}=\frac{1}{\sqrt{3}} \Rightarrow \mathrm{ae}=\frac{1}{\sqrt{6}}\)Foci \(\quad(\mathrm{h} \pm \mathrm{ae}, \mathrm{k}) \equiv\left(1 \pm \frac{1}{\sqrt{6}}, 2\right)\)
AP EAMCET-17.09.2020
Ellipse
120546
If the line joining the points \(A(\alpha)\) and \(B(\beta)\) on the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) is focal chord, then one possible value of \(\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2}\) is
1 -3
2 3
3 -9
4 9
Explanation:
C Given, ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\)
Here, \(\mathrm{a}=5\) and \(\mathrm{b}=3\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}\)
Equation of chord joining \(A(\alpha)\) and \(B(\beta)\) is
\(\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)\)
It passes through focus \((\mathrm{ae}, 0)\)
\(\frac{\mathrm{ae}}{\mathrm{a}} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{0}{\mathrm{~b}} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)\)
\(\frac{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}{\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{1}{\mathrm{e}}\)
\(\frac{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)+\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}{\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)-\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{1+\mathrm{e}}{1-\mathrm{e}}\)
\(\frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{-2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=\frac{1+\mathrm{e}}{1-\mathrm{e}}\)
\(\cot \left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right)=\frac{e+1}{e-1}=\frac{\frac{4}{5}+1}{\frac{4}{5}-1}=-9\)
\(\cot \left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right)=-9\)
