120517
Equation of the ellipse whose foci are \((2,2)\) and \((4,2)\) and the major axis is of length 10 , is:
1 \(\frac{(x+3)^2}{24}+\frac{(y+2)^2}{25}=1\)
2 \(\frac{(x-3)^2}{24}+\frac{(y-2)^2}{25}=1\)
3 \(\frac{(x+3)^2}{25}+\frac{(y+2)^2}{24}=1\)
4 \(\frac{(x-3)^2}{25}+\frac{(y-2)^2}{24}=1\)
Explanation:
D Given,
Focci of an ellipse are \((2,2)\) and \((4,2)\) and major axis is of length 10
\(\therefore \quad 2 \mathrm{ae}=2\)
\(2 \mathrm{a}=10\)
\(\mathrm{a}=5\)
\(\mathrm{a}^2=25\)
\(\therefore \quad 2 \times 5 \times \mathrm{e}=2 \mathrm{e}=\frac{1}{5}\)
\(\because \quad \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \mathrm{b}^2=25\left(1-\frac{1}{25}\right)\)
\(\mathrm{b}^2=24\)
we know that, centre of ellipse \(=\) mid point of foci
\(=\left(\frac{2+4}{2}, \frac{2+2}{2}\right)=(3,2)\)
Hence, equation of ellipse is
\(\frac{(\mathrm{x}-3)^2}{25}+\frac{(\mathrm{y}-2)^2}{24}=1\)
BCECE-2014]
Ellipse
120518
The eccentricity of an ellipse whose pair of a conjugate diameter are \(y=x\) and \(3 y=-2 x\) is
1 \(\frac{2}{3}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{\sqrt{3}}\)
4 None of these
Explanation:
C Given,
Conjugate diameter are \(y=x\) and \(3 y=-2 x\)
For conjugate diameter \(\left(\mathrm{m}_1 \times \mathrm{m}_2\right)=\frac{-\mathrm{b}^2}{\mathrm{a}^2}\)
Where \(\mathrm{m}_1\) and \(\mathrm{m}_2\) are slopes of the respective diameters
\(1 \times\left(\frac{-2}{3}\right)=\frac{-\mathrm{b}^2}{\mathrm{a}^2}\)
\(\mathrm{~b}^2=\frac{2}{3} \mathrm{a}^2\)
How, \(\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\frac{2}{3} \mathrm{a}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{2}{3}=1-\mathrm{e}^2\)
\(\mathrm{e}^2=\frac{1}{3}=\frac{1}{\sqrt{3}}\)
JCECE-2012
Ellipse
120519
The length of latusrectum of the ellipse \(2 x^2+y^2\) \(-8 \mathrm{x}+2 \mathrm{y}+7=0\), is
120520
The equation of ellipse whose axes are coincident with the coordinate axes and which touches the straight lines \(3 x-2 y-20=0\) and \(x\) \(+6 y-20=0\), is
1 \(\frac{x^2}{5}+\frac{y^2}{8}=1\)
2 \(\frac{x^2}{40}+\frac{y^2}{10}=10\)
3 \(\frac{x^2}{40}+\frac{y^2}{10}=1\)
4 \(\frac{x^2}{10}+\frac{y^2}{40}=1\)
Explanation:
C The equation of the ellipse be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\therefore\) The general equation of the tangent of the ellipse is
\(\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
Now,
\(3 \mathrm{x}-2 \mathrm{y}-20=0\)
\(\therefore \quad \mathrm{y}=\frac{3}{2} \mathrm{x}-10\) is a tangent to the ellipse
Therefore, on comparing with equation (i), we get-
\(\mathrm{m}=\frac{3}{2} \text { and } \quad \mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2=100\)
\(\mathrm{a}^2 \cdot \frac{9}{4}+\mathrm{b}^2=100\)
\(9 \mathrm{a}^2+4 \mathrm{~b}^2=400\)
Similarly, line \(x+6 y-20=0\)
\(\mathrm{y}=-\frac{1}{6} \mathrm{x}+\frac{10}{3}\)
is a tangent to the ellipse is therefore on comparing with equation(i), we get :-
\(\mathrm{m}=-\frac{1}{6} \text { and } \mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2=\frac{100}{9}\)
\(\frac{\mathrm{a}^2}{36}+\mathrm{b}^2=\frac{100}{9}\)
\(\mathrm{a}^2+36 \mathrm{~b}^2=400\)
From equation (ii) and (iii), we get-
\(\mathrm{a}^2=40 \text { and } \mathrm{b}^2=10\)Hence, \(\frac{x^2}{40}+\frac{y^2}{10}=1\) is required equations of ellipse.
120517
Equation of the ellipse whose foci are \((2,2)\) and \((4,2)\) and the major axis is of length 10 , is:
1 \(\frac{(x+3)^2}{24}+\frac{(y+2)^2}{25}=1\)
2 \(\frac{(x-3)^2}{24}+\frac{(y-2)^2}{25}=1\)
3 \(\frac{(x+3)^2}{25}+\frac{(y+2)^2}{24}=1\)
4 \(\frac{(x-3)^2}{25}+\frac{(y-2)^2}{24}=1\)
Explanation:
D Given,
Focci of an ellipse are \((2,2)\) and \((4,2)\) and major axis is of length 10
\(\therefore \quad 2 \mathrm{ae}=2\)
\(2 \mathrm{a}=10\)
\(\mathrm{a}=5\)
\(\mathrm{a}^2=25\)
\(\therefore \quad 2 \times 5 \times \mathrm{e}=2 \mathrm{e}=\frac{1}{5}\)
\(\because \quad \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \mathrm{b}^2=25\left(1-\frac{1}{25}\right)\)
\(\mathrm{b}^2=24\)
we know that, centre of ellipse \(=\) mid point of foci
\(=\left(\frac{2+4}{2}, \frac{2+2}{2}\right)=(3,2)\)
Hence, equation of ellipse is
\(\frac{(\mathrm{x}-3)^2}{25}+\frac{(\mathrm{y}-2)^2}{24}=1\)
BCECE-2014]
Ellipse
120518
The eccentricity of an ellipse whose pair of a conjugate diameter are \(y=x\) and \(3 y=-2 x\) is
1 \(\frac{2}{3}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{\sqrt{3}}\)
4 None of these
Explanation:
C Given,
Conjugate diameter are \(y=x\) and \(3 y=-2 x\)
For conjugate diameter \(\left(\mathrm{m}_1 \times \mathrm{m}_2\right)=\frac{-\mathrm{b}^2}{\mathrm{a}^2}\)
Where \(\mathrm{m}_1\) and \(\mathrm{m}_2\) are slopes of the respective diameters
\(1 \times\left(\frac{-2}{3}\right)=\frac{-\mathrm{b}^2}{\mathrm{a}^2}\)
\(\mathrm{~b}^2=\frac{2}{3} \mathrm{a}^2\)
How, \(\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\frac{2}{3} \mathrm{a}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{2}{3}=1-\mathrm{e}^2\)
\(\mathrm{e}^2=\frac{1}{3}=\frac{1}{\sqrt{3}}\)
JCECE-2012
Ellipse
120519
The length of latusrectum of the ellipse \(2 x^2+y^2\) \(-8 \mathrm{x}+2 \mathrm{y}+7=0\), is
120520
The equation of ellipse whose axes are coincident with the coordinate axes and which touches the straight lines \(3 x-2 y-20=0\) and \(x\) \(+6 y-20=0\), is
1 \(\frac{x^2}{5}+\frac{y^2}{8}=1\)
2 \(\frac{x^2}{40}+\frac{y^2}{10}=10\)
3 \(\frac{x^2}{40}+\frac{y^2}{10}=1\)
4 \(\frac{x^2}{10}+\frac{y^2}{40}=1\)
Explanation:
C The equation of the ellipse be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\therefore\) The general equation of the tangent of the ellipse is
\(\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
Now,
\(3 \mathrm{x}-2 \mathrm{y}-20=0\)
\(\therefore \quad \mathrm{y}=\frac{3}{2} \mathrm{x}-10\) is a tangent to the ellipse
Therefore, on comparing with equation (i), we get-
\(\mathrm{m}=\frac{3}{2} \text { and } \quad \mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2=100\)
\(\mathrm{a}^2 \cdot \frac{9}{4}+\mathrm{b}^2=100\)
\(9 \mathrm{a}^2+4 \mathrm{~b}^2=400\)
Similarly, line \(x+6 y-20=0\)
\(\mathrm{y}=-\frac{1}{6} \mathrm{x}+\frac{10}{3}\)
is a tangent to the ellipse is therefore on comparing with equation(i), we get :-
\(\mathrm{m}=-\frac{1}{6} \text { and } \mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2=\frac{100}{9}\)
\(\frac{\mathrm{a}^2}{36}+\mathrm{b}^2=\frac{100}{9}\)
\(\mathrm{a}^2+36 \mathrm{~b}^2=400\)
From equation (ii) and (iii), we get-
\(\mathrm{a}^2=40 \text { and } \mathrm{b}^2=10\)Hence, \(\frac{x^2}{40}+\frac{y^2}{10}=1\) is required equations of ellipse.
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Ellipse
120517
Equation of the ellipse whose foci are \((2,2)\) and \((4,2)\) and the major axis is of length 10 , is:
1 \(\frac{(x+3)^2}{24}+\frac{(y+2)^2}{25}=1\)
2 \(\frac{(x-3)^2}{24}+\frac{(y-2)^2}{25}=1\)
3 \(\frac{(x+3)^2}{25}+\frac{(y+2)^2}{24}=1\)
4 \(\frac{(x-3)^2}{25}+\frac{(y-2)^2}{24}=1\)
Explanation:
D Given,
Focci of an ellipse are \((2,2)\) and \((4,2)\) and major axis is of length 10
\(\therefore \quad 2 \mathrm{ae}=2\)
\(2 \mathrm{a}=10\)
\(\mathrm{a}=5\)
\(\mathrm{a}^2=25\)
\(\therefore \quad 2 \times 5 \times \mathrm{e}=2 \mathrm{e}=\frac{1}{5}\)
\(\because \quad \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \mathrm{b}^2=25\left(1-\frac{1}{25}\right)\)
\(\mathrm{b}^2=24\)
we know that, centre of ellipse \(=\) mid point of foci
\(=\left(\frac{2+4}{2}, \frac{2+2}{2}\right)=(3,2)\)
Hence, equation of ellipse is
\(\frac{(\mathrm{x}-3)^2}{25}+\frac{(\mathrm{y}-2)^2}{24}=1\)
BCECE-2014]
Ellipse
120518
The eccentricity of an ellipse whose pair of a conjugate diameter are \(y=x\) and \(3 y=-2 x\) is
1 \(\frac{2}{3}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{\sqrt{3}}\)
4 None of these
Explanation:
C Given,
Conjugate diameter are \(y=x\) and \(3 y=-2 x\)
For conjugate diameter \(\left(\mathrm{m}_1 \times \mathrm{m}_2\right)=\frac{-\mathrm{b}^2}{\mathrm{a}^2}\)
Where \(\mathrm{m}_1\) and \(\mathrm{m}_2\) are slopes of the respective diameters
\(1 \times\left(\frac{-2}{3}\right)=\frac{-\mathrm{b}^2}{\mathrm{a}^2}\)
\(\mathrm{~b}^2=\frac{2}{3} \mathrm{a}^2\)
How, \(\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\frac{2}{3} \mathrm{a}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{2}{3}=1-\mathrm{e}^2\)
\(\mathrm{e}^2=\frac{1}{3}=\frac{1}{\sqrt{3}}\)
JCECE-2012
Ellipse
120519
The length of latusrectum of the ellipse \(2 x^2+y^2\) \(-8 \mathrm{x}+2 \mathrm{y}+7=0\), is
120520
The equation of ellipse whose axes are coincident with the coordinate axes and which touches the straight lines \(3 x-2 y-20=0\) and \(x\) \(+6 y-20=0\), is
1 \(\frac{x^2}{5}+\frac{y^2}{8}=1\)
2 \(\frac{x^2}{40}+\frac{y^2}{10}=10\)
3 \(\frac{x^2}{40}+\frac{y^2}{10}=1\)
4 \(\frac{x^2}{10}+\frac{y^2}{40}=1\)
Explanation:
C The equation of the ellipse be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\therefore\) The general equation of the tangent of the ellipse is
\(\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
Now,
\(3 \mathrm{x}-2 \mathrm{y}-20=0\)
\(\therefore \quad \mathrm{y}=\frac{3}{2} \mathrm{x}-10\) is a tangent to the ellipse
Therefore, on comparing with equation (i), we get-
\(\mathrm{m}=\frac{3}{2} \text { and } \quad \mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2=100\)
\(\mathrm{a}^2 \cdot \frac{9}{4}+\mathrm{b}^2=100\)
\(9 \mathrm{a}^2+4 \mathrm{~b}^2=400\)
Similarly, line \(x+6 y-20=0\)
\(\mathrm{y}=-\frac{1}{6} \mathrm{x}+\frac{10}{3}\)
is a tangent to the ellipse is therefore on comparing with equation(i), we get :-
\(\mathrm{m}=-\frac{1}{6} \text { and } \mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2=\frac{100}{9}\)
\(\frac{\mathrm{a}^2}{36}+\mathrm{b}^2=\frac{100}{9}\)
\(\mathrm{a}^2+36 \mathrm{~b}^2=400\)
From equation (ii) and (iii), we get-
\(\mathrm{a}^2=40 \text { and } \mathrm{b}^2=10\)Hence, \(\frac{x^2}{40}+\frac{y^2}{10}=1\) is required equations of ellipse.
120517
Equation of the ellipse whose foci are \((2,2)\) and \((4,2)\) and the major axis is of length 10 , is:
1 \(\frac{(x+3)^2}{24}+\frac{(y+2)^2}{25}=1\)
2 \(\frac{(x-3)^2}{24}+\frac{(y-2)^2}{25}=1\)
3 \(\frac{(x+3)^2}{25}+\frac{(y+2)^2}{24}=1\)
4 \(\frac{(x-3)^2}{25}+\frac{(y-2)^2}{24}=1\)
Explanation:
D Given,
Focci of an ellipse are \((2,2)\) and \((4,2)\) and major axis is of length 10
\(\therefore \quad 2 \mathrm{ae}=2\)
\(2 \mathrm{a}=10\)
\(\mathrm{a}=5\)
\(\mathrm{a}^2=25\)
\(\therefore \quad 2 \times 5 \times \mathrm{e}=2 \mathrm{e}=\frac{1}{5}\)
\(\because \quad \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \mathrm{b}^2=25\left(1-\frac{1}{25}\right)\)
\(\mathrm{b}^2=24\)
we know that, centre of ellipse \(=\) mid point of foci
\(=\left(\frac{2+4}{2}, \frac{2+2}{2}\right)=(3,2)\)
Hence, equation of ellipse is
\(\frac{(\mathrm{x}-3)^2}{25}+\frac{(\mathrm{y}-2)^2}{24}=1\)
BCECE-2014]
Ellipse
120518
The eccentricity of an ellipse whose pair of a conjugate diameter are \(y=x\) and \(3 y=-2 x\) is
1 \(\frac{2}{3}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{\sqrt{3}}\)
4 None of these
Explanation:
C Given,
Conjugate diameter are \(y=x\) and \(3 y=-2 x\)
For conjugate diameter \(\left(\mathrm{m}_1 \times \mathrm{m}_2\right)=\frac{-\mathrm{b}^2}{\mathrm{a}^2}\)
Where \(\mathrm{m}_1\) and \(\mathrm{m}_2\) are slopes of the respective diameters
\(1 \times\left(\frac{-2}{3}\right)=\frac{-\mathrm{b}^2}{\mathrm{a}^2}\)
\(\mathrm{~b}^2=\frac{2}{3} \mathrm{a}^2\)
How, \(\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\frac{2}{3} \mathrm{a}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{2}{3}=1-\mathrm{e}^2\)
\(\mathrm{e}^2=\frac{1}{3}=\frac{1}{\sqrt{3}}\)
JCECE-2012
Ellipse
120519
The length of latusrectum of the ellipse \(2 x^2+y^2\) \(-8 \mathrm{x}+2 \mathrm{y}+7=0\), is
120520
The equation of ellipse whose axes are coincident with the coordinate axes and which touches the straight lines \(3 x-2 y-20=0\) and \(x\) \(+6 y-20=0\), is
1 \(\frac{x^2}{5}+\frac{y^2}{8}=1\)
2 \(\frac{x^2}{40}+\frac{y^2}{10}=10\)
3 \(\frac{x^2}{40}+\frac{y^2}{10}=1\)
4 \(\frac{x^2}{10}+\frac{y^2}{40}=1\)
Explanation:
C The equation of the ellipse be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\therefore\) The general equation of the tangent of the ellipse is
\(\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
Now,
\(3 \mathrm{x}-2 \mathrm{y}-20=0\)
\(\therefore \quad \mathrm{y}=\frac{3}{2} \mathrm{x}-10\) is a tangent to the ellipse
Therefore, on comparing with equation (i), we get-
\(\mathrm{m}=\frac{3}{2} \text { and } \quad \mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2=100\)
\(\mathrm{a}^2 \cdot \frac{9}{4}+\mathrm{b}^2=100\)
\(9 \mathrm{a}^2+4 \mathrm{~b}^2=400\)
Similarly, line \(x+6 y-20=0\)
\(\mathrm{y}=-\frac{1}{6} \mathrm{x}+\frac{10}{3}\)
is a tangent to the ellipse is therefore on comparing with equation(i), we get :-
\(\mathrm{m}=-\frac{1}{6} \text { and } \mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2=\frac{100}{9}\)
\(\frac{\mathrm{a}^2}{36}+\mathrm{b}^2=\frac{100}{9}\)
\(\mathrm{a}^2+36 \mathrm{~b}^2=400\)
From equation (ii) and (iii), we get-
\(\mathrm{a}^2=40 \text { and } \mathrm{b}^2=10\)Hence, \(\frac{x^2}{40}+\frac{y^2}{10}=1\) is required equations of ellipse.