120547
Let \(P\) is variable point on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with foci \(F_1\) and \(F_2\). If \(A\) is the area of the triangle \(P F_1 F_2\), then the maximum value of \(A\) is
1 \(\frac{\mathrm{e}}{\mathrm{ab}}\)
2 \(\frac{\mathrm{ae}}{\mathrm{b}}\)
3 aeb
4 \(\frac{\mathrm{ab}}{\mathrm{e}}\)
Explanation:
C Given, ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\mathrm{A}=\operatorname{Area} \Delta\left(\mathrm{PS}_1 \mathrm{~S}_2\right)\)
\(\mathrm{A}=\frac{1}{2}\left|\begin{array}{ccc}\mathrm{a} \cos \theta & \mathrm{b} \sin \theta & 1 \\ \mathrm{ae} & 0 & 1 \\ -\mathrm{ae} & 0 & 1\end{array}\right|\)
\(=\frac{1}{2} \mathrm{~b} \sin \theta \times 2 \mathrm{ae}=\mathrm{abe} \sin \theta\)
It is maximum when, \(\theta=\frac{\pi}{2}\)
\(\therefore\) Maximum value of \(\mathrm{A}=\) abe
AP EAMCET-20.04.2019
Ellipse
120548
The eccentric angle of a point on the ellipse \(x^2+\) \(3 y^2=6\) lying at a distance of 2 units from its centre is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, ellipse : \(x^2+3 y^2=6\)
\(\frac{x^2}{(\sqrt{6})^2}+\frac{y^2}{(\sqrt{2})^2}=1\)
Let, eccentric angle of the point be \(\theta\), then its coordinate \((\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)\).
\(\because\) Distance of point from the centre is 2 units
\(\therefore \quad(\sqrt{6} \cos \theta-0)^2+(\sqrt{2} \sin \theta-0)^2=4\)
\(6 \cos ^2 \theta+2\left(1-\cos ^2 \theta\right)=4 \quad\left(\because \sin ^2 \theta=1-\cos ^2 \theta\right)\)
\(4 \cos ^2 \theta=2\)
\(\quad \cos \theta= \pm \frac{1}{\sqrt{2}}\)
\(\therefore \quad \theta=\frac{\pi}{4} \text { or } \frac{3 \pi}{4}\)
AP EAMCET-04.07.2021
Ellipse
120549
The eccentricity of an ellipse, with its centre as origin, is \(1 / 2\). If one of the directrices is \(x=4\), then the equation of the ellipse is given by
1 \(4 x^2+y^2=12\)
2 \(x^2+3 y^2=12\)
3 \(4 x^2+3 y^2=12\)
4 \(3 x^2+4 y^2=12\)
Explanation:
D Let the equation of ellipse, \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Here, \(b^2=a^2\left(1-e^2\right)\)
Given, \(\mathrm{e}=\frac{1}{2}\)
\(\mathrm{b}^2=\frac{3}{4}\left(\mathrm{a}^2\right)\)
Also, \(\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\mathrm{a}=2\)
Putting the value of ' \(a\) ' in equation (i), we get\(\mathrm{b}=\sqrt{3}\)
Hence, the required equation,
\(\frac{x^2}{4}+\frac{y^2}{3}=1\)
\(y^2=12\)\(3 \mathrm{x}^2+4 \mathrm{y}^2=12\)
AP EAMCET-18.09.2020
Ellipse
120550
The equation of the ellipse with its focus at (6, 2) centre at \((1,2)\) and which passes through the point \((4,6)\) is
1 \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{16}=1\)
2 \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{20}=1\)
3 \(\frac{(x-1)^2}{45}+\frac{(y-1)^2}{16}=1\)
4 \(\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1\)
Explanation:
D Given that, focus \((6,2)\) and centre \((1,2)\)
Equation of ellipse : \(\frac{(x-1)^2}{a^2}+\frac{(y-2)^2}{b^2}=1\)
Distance between focus and centre is ' \(\mathrm{d}\) '.
So, \(\quad d=\sqrt{(1-6)^2+(2-2)^2}\)
\(\mathrm{d}=5\)
We have, \(\mathrm{a}^2-\mathrm{b}^2=25\)
\((4,6)\) lies on ellipse
\(\frac{(4-1)^2}{a^2}+\frac{(6-2)^2}{b^2}=1\)
\(9 b^2+16 a^2=a^2 b^2\)
From equation (i) and (ii), we get-
\(9 b^2+400+16 b^2=25 b^2+b^2\)
\(b^4=400\)
\(\left(b^2\right)^2-(20)^2=0\)
\(\left(b^2-20\right)\left(b^2+20\right)=0\)
\(b^2=20 \text { and } b^2=-20\)
Putting the value of ' \(b^2\) ' in equation (i), we get -
\(\mathrm{a}^2=45\)
Hence, the required equation -
\(\frac{(\mathrm{x}-1)^2}{45}+\frac{(\mathrm{y}-2)^2}{20}=1\)
120547
Let \(P\) is variable point on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with foci \(F_1\) and \(F_2\). If \(A\) is the area of the triangle \(P F_1 F_2\), then the maximum value of \(A\) is
1 \(\frac{\mathrm{e}}{\mathrm{ab}}\)
2 \(\frac{\mathrm{ae}}{\mathrm{b}}\)
3 aeb
4 \(\frac{\mathrm{ab}}{\mathrm{e}}\)
Explanation:
C Given, ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\mathrm{A}=\operatorname{Area} \Delta\left(\mathrm{PS}_1 \mathrm{~S}_2\right)\)
\(\mathrm{A}=\frac{1}{2}\left|\begin{array}{ccc}\mathrm{a} \cos \theta & \mathrm{b} \sin \theta & 1 \\ \mathrm{ae} & 0 & 1 \\ -\mathrm{ae} & 0 & 1\end{array}\right|\)
\(=\frac{1}{2} \mathrm{~b} \sin \theta \times 2 \mathrm{ae}=\mathrm{abe} \sin \theta\)
It is maximum when, \(\theta=\frac{\pi}{2}\)
\(\therefore\) Maximum value of \(\mathrm{A}=\) abe
AP EAMCET-20.04.2019
Ellipse
120548
The eccentric angle of a point on the ellipse \(x^2+\) \(3 y^2=6\) lying at a distance of 2 units from its centre is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, ellipse : \(x^2+3 y^2=6\)
\(\frac{x^2}{(\sqrt{6})^2}+\frac{y^2}{(\sqrt{2})^2}=1\)
Let, eccentric angle of the point be \(\theta\), then its coordinate \((\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)\).
\(\because\) Distance of point from the centre is 2 units
\(\therefore \quad(\sqrt{6} \cos \theta-0)^2+(\sqrt{2} \sin \theta-0)^2=4\)
\(6 \cos ^2 \theta+2\left(1-\cos ^2 \theta\right)=4 \quad\left(\because \sin ^2 \theta=1-\cos ^2 \theta\right)\)
\(4 \cos ^2 \theta=2\)
\(\quad \cos \theta= \pm \frac{1}{\sqrt{2}}\)
\(\therefore \quad \theta=\frac{\pi}{4} \text { or } \frac{3 \pi}{4}\)
AP EAMCET-04.07.2021
Ellipse
120549
The eccentricity of an ellipse, with its centre as origin, is \(1 / 2\). If one of the directrices is \(x=4\), then the equation of the ellipse is given by
1 \(4 x^2+y^2=12\)
2 \(x^2+3 y^2=12\)
3 \(4 x^2+3 y^2=12\)
4 \(3 x^2+4 y^2=12\)
Explanation:
D Let the equation of ellipse, \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Here, \(b^2=a^2\left(1-e^2\right)\)
Given, \(\mathrm{e}=\frac{1}{2}\)
\(\mathrm{b}^2=\frac{3}{4}\left(\mathrm{a}^2\right)\)
Also, \(\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\mathrm{a}=2\)
Putting the value of ' \(a\) ' in equation (i), we get\(\mathrm{b}=\sqrt{3}\)
Hence, the required equation,
\(\frac{x^2}{4}+\frac{y^2}{3}=1\)
\(y^2=12\)\(3 \mathrm{x}^2+4 \mathrm{y}^2=12\)
AP EAMCET-18.09.2020
Ellipse
120550
The equation of the ellipse with its focus at (6, 2) centre at \((1,2)\) and which passes through the point \((4,6)\) is
1 \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{16}=1\)
2 \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{20}=1\)
3 \(\frac{(x-1)^2}{45}+\frac{(y-1)^2}{16}=1\)
4 \(\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1\)
Explanation:
D Given that, focus \((6,2)\) and centre \((1,2)\)
Equation of ellipse : \(\frac{(x-1)^2}{a^2}+\frac{(y-2)^2}{b^2}=1\)
Distance between focus and centre is ' \(\mathrm{d}\) '.
So, \(\quad d=\sqrt{(1-6)^2+(2-2)^2}\)
\(\mathrm{d}=5\)
We have, \(\mathrm{a}^2-\mathrm{b}^2=25\)
\((4,6)\) lies on ellipse
\(\frac{(4-1)^2}{a^2}+\frac{(6-2)^2}{b^2}=1\)
\(9 b^2+16 a^2=a^2 b^2\)
From equation (i) and (ii), we get-
\(9 b^2+400+16 b^2=25 b^2+b^2\)
\(b^4=400\)
\(\left(b^2\right)^2-(20)^2=0\)
\(\left(b^2-20\right)\left(b^2+20\right)=0\)
\(b^2=20 \text { and } b^2=-20\)
Putting the value of ' \(b^2\) ' in equation (i), we get -
\(\mathrm{a}^2=45\)
Hence, the required equation -
\(\frac{(\mathrm{x}-1)^2}{45}+\frac{(\mathrm{y}-2)^2}{20}=1\)
120547
Let \(P\) is variable point on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with foci \(F_1\) and \(F_2\). If \(A\) is the area of the triangle \(P F_1 F_2\), then the maximum value of \(A\) is
1 \(\frac{\mathrm{e}}{\mathrm{ab}}\)
2 \(\frac{\mathrm{ae}}{\mathrm{b}}\)
3 aeb
4 \(\frac{\mathrm{ab}}{\mathrm{e}}\)
Explanation:
C Given, ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\mathrm{A}=\operatorname{Area} \Delta\left(\mathrm{PS}_1 \mathrm{~S}_2\right)\)
\(\mathrm{A}=\frac{1}{2}\left|\begin{array}{ccc}\mathrm{a} \cos \theta & \mathrm{b} \sin \theta & 1 \\ \mathrm{ae} & 0 & 1 \\ -\mathrm{ae} & 0 & 1\end{array}\right|\)
\(=\frac{1}{2} \mathrm{~b} \sin \theta \times 2 \mathrm{ae}=\mathrm{abe} \sin \theta\)
It is maximum when, \(\theta=\frac{\pi}{2}\)
\(\therefore\) Maximum value of \(\mathrm{A}=\) abe
AP EAMCET-20.04.2019
Ellipse
120548
The eccentric angle of a point on the ellipse \(x^2+\) \(3 y^2=6\) lying at a distance of 2 units from its centre is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, ellipse : \(x^2+3 y^2=6\)
\(\frac{x^2}{(\sqrt{6})^2}+\frac{y^2}{(\sqrt{2})^2}=1\)
Let, eccentric angle of the point be \(\theta\), then its coordinate \((\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)\).
\(\because\) Distance of point from the centre is 2 units
\(\therefore \quad(\sqrt{6} \cos \theta-0)^2+(\sqrt{2} \sin \theta-0)^2=4\)
\(6 \cos ^2 \theta+2\left(1-\cos ^2 \theta\right)=4 \quad\left(\because \sin ^2 \theta=1-\cos ^2 \theta\right)\)
\(4 \cos ^2 \theta=2\)
\(\quad \cos \theta= \pm \frac{1}{\sqrt{2}}\)
\(\therefore \quad \theta=\frac{\pi}{4} \text { or } \frac{3 \pi}{4}\)
AP EAMCET-04.07.2021
Ellipse
120549
The eccentricity of an ellipse, with its centre as origin, is \(1 / 2\). If one of the directrices is \(x=4\), then the equation of the ellipse is given by
1 \(4 x^2+y^2=12\)
2 \(x^2+3 y^2=12\)
3 \(4 x^2+3 y^2=12\)
4 \(3 x^2+4 y^2=12\)
Explanation:
D Let the equation of ellipse, \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Here, \(b^2=a^2\left(1-e^2\right)\)
Given, \(\mathrm{e}=\frac{1}{2}\)
\(\mathrm{b}^2=\frac{3}{4}\left(\mathrm{a}^2\right)\)
Also, \(\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\mathrm{a}=2\)
Putting the value of ' \(a\) ' in equation (i), we get\(\mathrm{b}=\sqrt{3}\)
Hence, the required equation,
\(\frac{x^2}{4}+\frac{y^2}{3}=1\)
\(y^2=12\)\(3 \mathrm{x}^2+4 \mathrm{y}^2=12\)
AP EAMCET-18.09.2020
Ellipse
120550
The equation of the ellipse with its focus at (6, 2) centre at \((1,2)\) and which passes through the point \((4,6)\) is
1 \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{16}=1\)
2 \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{20}=1\)
3 \(\frac{(x-1)^2}{45}+\frac{(y-1)^2}{16}=1\)
4 \(\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1\)
Explanation:
D Given that, focus \((6,2)\) and centre \((1,2)\)
Equation of ellipse : \(\frac{(x-1)^2}{a^2}+\frac{(y-2)^2}{b^2}=1\)
Distance between focus and centre is ' \(\mathrm{d}\) '.
So, \(\quad d=\sqrt{(1-6)^2+(2-2)^2}\)
\(\mathrm{d}=5\)
We have, \(\mathrm{a}^2-\mathrm{b}^2=25\)
\((4,6)\) lies on ellipse
\(\frac{(4-1)^2}{a^2}+\frac{(6-2)^2}{b^2}=1\)
\(9 b^2+16 a^2=a^2 b^2\)
From equation (i) and (ii), we get-
\(9 b^2+400+16 b^2=25 b^2+b^2\)
\(b^4=400\)
\(\left(b^2\right)^2-(20)^2=0\)
\(\left(b^2-20\right)\left(b^2+20\right)=0\)
\(b^2=20 \text { and } b^2=-20\)
Putting the value of ' \(b^2\) ' in equation (i), we get -
\(\mathrm{a}^2=45\)
Hence, the required equation -
\(\frac{(\mathrm{x}-1)^2}{45}+\frac{(\mathrm{y}-2)^2}{20}=1\)
120547
Let \(P\) is variable point on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with foci \(F_1\) and \(F_2\). If \(A\) is the area of the triangle \(P F_1 F_2\), then the maximum value of \(A\) is
1 \(\frac{\mathrm{e}}{\mathrm{ab}}\)
2 \(\frac{\mathrm{ae}}{\mathrm{b}}\)
3 aeb
4 \(\frac{\mathrm{ab}}{\mathrm{e}}\)
Explanation:
C Given, ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\mathrm{A}=\operatorname{Area} \Delta\left(\mathrm{PS}_1 \mathrm{~S}_2\right)\)
\(\mathrm{A}=\frac{1}{2}\left|\begin{array}{ccc}\mathrm{a} \cos \theta & \mathrm{b} \sin \theta & 1 \\ \mathrm{ae} & 0 & 1 \\ -\mathrm{ae} & 0 & 1\end{array}\right|\)
\(=\frac{1}{2} \mathrm{~b} \sin \theta \times 2 \mathrm{ae}=\mathrm{abe} \sin \theta\)
It is maximum when, \(\theta=\frac{\pi}{2}\)
\(\therefore\) Maximum value of \(\mathrm{A}=\) abe
AP EAMCET-20.04.2019
Ellipse
120548
The eccentric angle of a point on the ellipse \(x^2+\) \(3 y^2=6\) lying at a distance of 2 units from its centre is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
B Given, ellipse : \(x^2+3 y^2=6\)
\(\frac{x^2}{(\sqrt{6})^2}+\frac{y^2}{(\sqrt{2})^2}=1\)
Let, eccentric angle of the point be \(\theta\), then its coordinate \((\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)\).
\(\because\) Distance of point from the centre is 2 units
\(\therefore \quad(\sqrt{6} \cos \theta-0)^2+(\sqrt{2} \sin \theta-0)^2=4\)
\(6 \cos ^2 \theta+2\left(1-\cos ^2 \theta\right)=4 \quad\left(\because \sin ^2 \theta=1-\cos ^2 \theta\right)\)
\(4 \cos ^2 \theta=2\)
\(\quad \cos \theta= \pm \frac{1}{\sqrt{2}}\)
\(\therefore \quad \theta=\frac{\pi}{4} \text { or } \frac{3 \pi}{4}\)
AP EAMCET-04.07.2021
Ellipse
120549
The eccentricity of an ellipse, with its centre as origin, is \(1 / 2\). If one of the directrices is \(x=4\), then the equation of the ellipse is given by
1 \(4 x^2+y^2=12\)
2 \(x^2+3 y^2=12\)
3 \(4 x^2+3 y^2=12\)
4 \(3 x^2+4 y^2=12\)
Explanation:
D Let the equation of ellipse, \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Here, \(b^2=a^2\left(1-e^2\right)\)
Given, \(\mathrm{e}=\frac{1}{2}\)
\(\mathrm{b}^2=\frac{3}{4}\left(\mathrm{a}^2\right)\)
Also, \(\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\mathrm{a}=2\)
Putting the value of ' \(a\) ' in equation (i), we get\(\mathrm{b}=\sqrt{3}\)
Hence, the required equation,
\(\frac{x^2}{4}+\frac{y^2}{3}=1\)
\(y^2=12\)\(3 \mathrm{x}^2+4 \mathrm{y}^2=12\)
AP EAMCET-18.09.2020
Ellipse
120550
The equation of the ellipse with its focus at (6, 2) centre at \((1,2)\) and which passes through the point \((4,6)\) is
1 \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{16}=1\)
2 \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{20}=1\)
3 \(\frac{(x-1)^2}{45}+\frac{(y-1)^2}{16}=1\)
4 \(\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1\)
Explanation:
D Given that, focus \((6,2)\) and centre \((1,2)\)
Equation of ellipse : \(\frac{(x-1)^2}{a^2}+\frac{(y-2)^2}{b^2}=1\)
Distance between focus and centre is ' \(\mathrm{d}\) '.
So, \(\quad d=\sqrt{(1-6)^2+(2-2)^2}\)
\(\mathrm{d}=5\)
We have, \(\mathrm{a}^2-\mathrm{b}^2=25\)
\((4,6)\) lies on ellipse
\(\frac{(4-1)^2}{a^2}+\frac{(6-2)^2}{b^2}=1\)
\(9 b^2+16 a^2=a^2 b^2\)
From equation (i) and (ii), we get-
\(9 b^2+400+16 b^2=25 b^2+b^2\)
\(b^4=400\)
\(\left(b^2\right)^2-(20)^2=0\)
\(\left(b^2-20\right)\left(b^2+20\right)=0\)
\(b^2=20 \text { and } b^2=-20\)
Putting the value of ' \(b^2\) ' in equation (i), we get -
\(\mathrm{a}^2=45\)
Hence, the required equation -
\(\frac{(\mathrm{x}-1)^2}{45}+\frac{(\mathrm{y}-2)^2}{20}=1\)
