120568
An ellipse having the coordinate axes as its axes and its major axes along \(Y\)-axis, passes through the point \((-3,1)\) and has eccentricity \(\sqrt{\frac{2}{5}}\). Then its equation is
1 \(3 x^2+5 y^2-15=0\)
2 \(5 x^2+3 y^2-32=0\)
3 \(3 x^2+5 y^2-32=0\)
4 \(5 \mathrm{x}^2+3 \mathrm{y}^2-48=0\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{\frac{2}{5}}\)
\(\sqrt{1-\frac{b^2}{a^2}}=\sqrt{\frac{2}{5}}\)
\(5\left(a^2-b^2\right)=2 a^2\)
\(a^2=\frac{5}{3} b^2\)
Equation of ellipse with point \((-3,1)\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1\)
\(\frac{9}{a^2}+\frac{1}{b^2}=1\)
\(9 b^2+a^2=a^2 b^2\)
\(9 b^2+\frac{5 b^2}{3}=\frac{5}{3} b^4 \quad \text { \{from (i)\} }\)
\(\frac{27 b^2+5 b^2}{3}=\frac{5}{3} b^4\)
\(\frac{5}{3} b^2=\frac{32}{3}\)
\(b^2=\frac{32}{5} \text { and } a^2=\frac{32}{3}\)
\{from (i) \(\}\)
Therefore, the equation of ellipse -
\(\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1\)
\(\frac{3 x^2}{32}+\frac{5 y^2}{32}=1\)
\(3 x^2+5 y^2-32=0\)
AP EAMCET-22.04.2018
Ellipse
120569
The line \(x=m\) meets an ellipse \(9 x^2+y^2=9\) in the real and distinct points if and only if
1 \(|\mathrm{m}|>1\)
2 \(|\mathrm{m}|\lt 1\)
3 \(|\mathrm{m}|>2\)
4 \(|\mathrm{m}|\lt 2\)
Explanation:
B Given,
\(9 x^2+y^2=9\)
\(\frac{x^2}{1}+\frac{y^2}{9}=1\)
Also, \(\mathrm{x}=\mathrm{m} \quad\) (put in \(\left.\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\)
So, \(\quad \mathrm{m}^2+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{y}^2=9\left(1-\mathrm{m}^2\right)\)
It gives real and distinct if,
\(1-\mathrm{m}^2>0\)
\(\mathrm{~m}^2\lt 1\)
\(|\mathrm{~m}|\lt 1\)
AP EAMCET-21.09.2020
Ellipse
120570
\(S\) and \(T\) are the foci of an ellipse and \(B\) is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Equation of the ellipse is -
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Foci are \(\mathrm{S}(\mathrm{ae}, 0) \mathrm{T}(-\mathrm{ae}, 0)\) \(\mathrm{B}(0, b)\) is the end of the minor axis.
\(\mathrm{SB}=\mathrm{ST}\)
\(\mathrm{SB}^2=\mathrm{ST}^2\)
\(\mathrm{~b}^2=3 \mathrm{a}^2 \mathrm{e}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=3 \mathrm{a}^2 \mathrm{e}^2\)
\(1-\mathrm{e}^2=3 \mathrm{e}^2\)
\(4 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{4}\)
\(\mathrm{e}=\frac{1}{2}\)Eccentricity of the ellipse, \(\mathrm{e}=\frac{1}{2}\)
WB JEE-2019
Ellipse
120571
If \(B\) and \(B\) ' are the ends of minor axis and \(S\) and \(S^{\prime}\) are the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) then the area of the rhombus SBS'B' will be
1 12 sq units
2 48 sq units
3 24 sq units
4 36 sq units
Explanation:
C Given,
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{25}}\)
\(\mathrm{e}=\frac{4}{5}\)
\(\therefore \mathrm{S}=(\mathrm{ae}, 0)=\left(5 \times \frac{4}{5}, 0\right)\)
\(\mathrm{S}=(4,0)\)
Area of rhombus \(=4 \times\) area of \(\Delta \mathrm{BOS}\),
\(=4 \times\left(\frac{1}{2} \times \mathrm{OS} \times \mathrm{OB}\right)\)
\(=4 \times \frac{1}{2} \times 4 \times 3\)
\(\mathrm{~A} =24 \text { sq. unit }\)
WB JEE-2020
Ellipse
120572
The equation \(\operatorname{rcos}\left(\theta-\frac{\pi}{3}\right)=2\) represents
1 a circle
2 a parabola
3 an ellipse
4 a straight line
Explanation:
D Given,
\(r \cos \left(\theta-\frac{\pi}{3}\right)=2\)
\(r\left[\cos \theta \cos \frac{\pi}{3}+\sin \theta \sin \frac{\pi}{3}\right]=2\)
\(\mathrm{r} \cos \theta \cdot \frac{1}{2}+\mathrm{r} \sin \theta \cdot \frac{\sqrt{3}}{2}=2\)
Let, \(\mathrm{r} \cos \theta=\mathrm{x}, \mathrm{y}=\mathrm{r} \sin \theta\)
Therefore, \(x+\sqrt{3} y=4\)
The equation is represents a straight line
120568
An ellipse having the coordinate axes as its axes and its major axes along \(Y\)-axis, passes through the point \((-3,1)\) and has eccentricity \(\sqrt{\frac{2}{5}}\). Then its equation is
1 \(3 x^2+5 y^2-15=0\)
2 \(5 x^2+3 y^2-32=0\)
3 \(3 x^2+5 y^2-32=0\)
4 \(5 \mathrm{x}^2+3 \mathrm{y}^2-48=0\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{\frac{2}{5}}\)
\(\sqrt{1-\frac{b^2}{a^2}}=\sqrt{\frac{2}{5}}\)
\(5\left(a^2-b^2\right)=2 a^2\)
\(a^2=\frac{5}{3} b^2\)
Equation of ellipse with point \((-3,1)\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1\)
\(\frac{9}{a^2}+\frac{1}{b^2}=1\)
\(9 b^2+a^2=a^2 b^2\)
\(9 b^2+\frac{5 b^2}{3}=\frac{5}{3} b^4 \quad \text { \{from (i)\} }\)
\(\frac{27 b^2+5 b^2}{3}=\frac{5}{3} b^4\)
\(\frac{5}{3} b^2=\frac{32}{3}\)
\(b^2=\frac{32}{5} \text { and } a^2=\frac{32}{3}\)
\{from (i) \(\}\)
Therefore, the equation of ellipse -
\(\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1\)
\(\frac{3 x^2}{32}+\frac{5 y^2}{32}=1\)
\(3 x^2+5 y^2-32=0\)
AP EAMCET-22.04.2018
Ellipse
120569
The line \(x=m\) meets an ellipse \(9 x^2+y^2=9\) in the real and distinct points if and only if
1 \(|\mathrm{m}|>1\)
2 \(|\mathrm{m}|\lt 1\)
3 \(|\mathrm{m}|>2\)
4 \(|\mathrm{m}|\lt 2\)
Explanation:
B Given,
\(9 x^2+y^2=9\)
\(\frac{x^2}{1}+\frac{y^2}{9}=1\)
Also, \(\mathrm{x}=\mathrm{m} \quad\) (put in \(\left.\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\)
So, \(\quad \mathrm{m}^2+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{y}^2=9\left(1-\mathrm{m}^2\right)\)
It gives real and distinct if,
\(1-\mathrm{m}^2>0\)
\(\mathrm{~m}^2\lt 1\)
\(|\mathrm{~m}|\lt 1\)
AP EAMCET-21.09.2020
Ellipse
120570
\(S\) and \(T\) are the foci of an ellipse and \(B\) is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Equation of the ellipse is -
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Foci are \(\mathrm{S}(\mathrm{ae}, 0) \mathrm{T}(-\mathrm{ae}, 0)\) \(\mathrm{B}(0, b)\) is the end of the minor axis.
\(\mathrm{SB}=\mathrm{ST}\)
\(\mathrm{SB}^2=\mathrm{ST}^2\)
\(\mathrm{~b}^2=3 \mathrm{a}^2 \mathrm{e}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=3 \mathrm{a}^2 \mathrm{e}^2\)
\(1-\mathrm{e}^2=3 \mathrm{e}^2\)
\(4 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{4}\)
\(\mathrm{e}=\frac{1}{2}\)Eccentricity of the ellipse, \(\mathrm{e}=\frac{1}{2}\)
WB JEE-2019
Ellipse
120571
If \(B\) and \(B\) ' are the ends of minor axis and \(S\) and \(S^{\prime}\) are the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) then the area of the rhombus SBS'B' will be
1 12 sq units
2 48 sq units
3 24 sq units
4 36 sq units
Explanation:
C Given,
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{25}}\)
\(\mathrm{e}=\frac{4}{5}\)
\(\therefore \mathrm{S}=(\mathrm{ae}, 0)=\left(5 \times \frac{4}{5}, 0\right)\)
\(\mathrm{S}=(4,0)\)
Area of rhombus \(=4 \times\) area of \(\Delta \mathrm{BOS}\),
\(=4 \times\left(\frac{1}{2} \times \mathrm{OS} \times \mathrm{OB}\right)\)
\(=4 \times \frac{1}{2} \times 4 \times 3\)
\(\mathrm{~A} =24 \text { sq. unit }\)
WB JEE-2020
Ellipse
120572
The equation \(\operatorname{rcos}\left(\theta-\frac{\pi}{3}\right)=2\) represents
1 a circle
2 a parabola
3 an ellipse
4 a straight line
Explanation:
D Given,
\(r \cos \left(\theta-\frac{\pi}{3}\right)=2\)
\(r\left[\cos \theta \cos \frac{\pi}{3}+\sin \theta \sin \frac{\pi}{3}\right]=2\)
\(\mathrm{r} \cos \theta \cdot \frac{1}{2}+\mathrm{r} \sin \theta \cdot \frac{\sqrt{3}}{2}=2\)
Let, \(\mathrm{r} \cos \theta=\mathrm{x}, \mathrm{y}=\mathrm{r} \sin \theta\)
Therefore, \(x+\sqrt{3} y=4\)
The equation is represents a straight line
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ellipse
120568
An ellipse having the coordinate axes as its axes and its major axes along \(Y\)-axis, passes through the point \((-3,1)\) and has eccentricity \(\sqrt{\frac{2}{5}}\). Then its equation is
1 \(3 x^2+5 y^2-15=0\)
2 \(5 x^2+3 y^2-32=0\)
3 \(3 x^2+5 y^2-32=0\)
4 \(5 \mathrm{x}^2+3 \mathrm{y}^2-48=0\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{\frac{2}{5}}\)
\(\sqrt{1-\frac{b^2}{a^2}}=\sqrt{\frac{2}{5}}\)
\(5\left(a^2-b^2\right)=2 a^2\)
\(a^2=\frac{5}{3} b^2\)
Equation of ellipse with point \((-3,1)\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1\)
\(\frac{9}{a^2}+\frac{1}{b^2}=1\)
\(9 b^2+a^2=a^2 b^2\)
\(9 b^2+\frac{5 b^2}{3}=\frac{5}{3} b^4 \quad \text { \{from (i)\} }\)
\(\frac{27 b^2+5 b^2}{3}=\frac{5}{3} b^4\)
\(\frac{5}{3} b^2=\frac{32}{3}\)
\(b^2=\frac{32}{5} \text { and } a^2=\frac{32}{3}\)
\{from (i) \(\}\)
Therefore, the equation of ellipse -
\(\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1\)
\(\frac{3 x^2}{32}+\frac{5 y^2}{32}=1\)
\(3 x^2+5 y^2-32=0\)
AP EAMCET-22.04.2018
Ellipse
120569
The line \(x=m\) meets an ellipse \(9 x^2+y^2=9\) in the real and distinct points if and only if
1 \(|\mathrm{m}|>1\)
2 \(|\mathrm{m}|\lt 1\)
3 \(|\mathrm{m}|>2\)
4 \(|\mathrm{m}|\lt 2\)
Explanation:
B Given,
\(9 x^2+y^2=9\)
\(\frac{x^2}{1}+\frac{y^2}{9}=1\)
Also, \(\mathrm{x}=\mathrm{m} \quad\) (put in \(\left.\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\)
So, \(\quad \mathrm{m}^2+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{y}^2=9\left(1-\mathrm{m}^2\right)\)
It gives real and distinct if,
\(1-\mathrm{m}^2>0\)
\(\mathrm{~m}^2\lt 1\)
\(|\mathrm{~m}|\lt 1\)
AP EAMCET-21.09.2020
Ellipse
120570
\(S\) and \(T\) are the foci of an ellipse and \(B\) is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Equation of the ellipse is -
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Foci are \(\mathrm{S}(\mathrm{ae}, 0) \mathrm{T}(-\mathrm{ae}, 0)\) \(\mathrm{B}(0, b)\) is the end of the minor axis.
\(\mathrm{SB}=\mathrm{ST}\)
\(\mathrm{SB}^2=\mathrm{ST}^2\)
\(\mathrm{~b}^2=3 \mathrm{a}^2 \mathrm{e}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=3 \mathrm{a}^2 \mathrm{e}^2\)
\(1-\mathrm{e}^2=3 \mathrm{e}^2\)
\(4 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{4}\)
\(\mathrm{e}=\frac{1}{2}\)Eccentricity of the ellipse, \(\mathrm{e}=\frac{1}{2}\)
WB JEE-2019
Ellipse
120571
If \(B\) and \(B\) ' are the ends of minor axis and \(S\) and \(S^{\prime}\) are the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) then the area of the rhombus SBS'B' will be
1 12 sq units
2 48 sq units
3 24 sq units
4 36 sq units
Explanation:
C Given,
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{25}}\)
\(\mathrm{e}=\frac{4}{5}\)
\(\therefore \mathrm{S}=(\mathrm{ae}, 0)=\left(5 \times \frac{4}{5}, 0\right)\)
\(\mathrm{S}=(4,0)\)
Area of rhombus \(=4 \times\) area of \(\Delta \mathrm{BOS}\),
\(=4 \times\left(\frac{1}{2} \times \mathrm{OS} \times \mathrm{OB}\right)\)
\(=4 \times \frac{1}{2} \times 4 \times 3\)
\(\mathrm{~A} =24 \text { sq. unit }\)
WB JEE-2020
Ellipse
120572
The equation \(\operatorname{rcos}\left(\theta-\frac{\pi}{3}\right)=2\) represents
1 a circle
2 a parabola
3 an ellipse
4 a straight line
Explanation:
D Given,
\(r \cos \left(\theta-\frac{\pi}{3}\right)=2\)
\(r\left[\cos \theta \cos \frac{\pi}{3}+\sin \theta \sin \frac{\pi}{3}\right]=2\)
\(\mathrm{r} \cos \theta \cdot \frac{1}{2}+\mathrm{r} \sin \theta \cdot \frac{\sqrt{3}}{2}=2\)
Let, \(\mathrm{r} \cos \theta=\mathrm{x}, \mathrm{y}=\mathrm{r} \sin \theta\)
Therefore, \(x+\sqrt{3} y=4\)
The equation is represents a straight line
120568
An ellipse having the coordinate axes as its axes and its major axes along \(Y\)-axis, passes through the point \((-3,1)\) and has eccentricity \(\sqrt{\frac{2}{5}}\). Then its equation is
1 \(3 x^2+5 y^2-15=0\)
2 \(5 x^2+3 y^2-32=0\)
3 \(3 x^2+5 y^2-32=0\)
4 \(5 \mathrm{x}^2+3 \mathrm{y}^2-48=0\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{\frac{2}{5}}\)
\(\sqrt{1-\frac{b^2}{a^2}}=\sqrt{\frac{2}{5}}\)
\(5\left(a^2-b^2\right)=2 a^2\)
\(a^2=\frac{5}{3} b^2\)
Equation of ellipse with point \((-3,1)\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1\)
\(\frac{9}{a^2}+\frac{1}{b^2}=1\)
\(9 b^2+a^2=a^2 b^2\)
\(9 b^2+\frac{5 b^2}{3}=\frac{5}{3} b^4 \quad \text { \{from (i)\} }\)
\(\frac{27 b^2+5 b^2}{3}=\frac{5}{3} b^4\)
\(\frac{5}{3} b^2=\frac{32}{3}\)
\(b^2=\frac{32}{5} \text { and } a^2=\frac{32}{3}\)
\{from (i) \(\}\)
Therefore, the equation of ellipse -
\(\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1\)
\(\frac{3 x^2}{32}+\frac{5 y^2}{32}=1\)
\(3 x^2+5 y^2-32=0\)
AP EAMCET-22.04.2018
Ellipse
120569
The line \(x=m\) meets an ellipse \(9 x^2+y^2=9\) in the real and distinct points if and only if
1 \(|\mathrm{m}|>1\)
2 \(|\mathrm{m}|\lt 1\)
3 \(|\mathrm{m}|>2\)
4 \(|\mathrm{m}|\lt 2\)
Explanation:
B Given,
\(9 x^2+y^2=9\)
\(\frac{x^2}{1}+\frac{y^2}{9}=1\)
Also, \(\mathrm{x}=\mathrm{m} \quad\) (put in \(\left.\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\)
So, \(\quad \mathrm{m}^2+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{y}^2=9\left(1-\mathrm{m}^2\right)\)
It gives real and distinct if,
\(1-\mathrm{m}^2>0\)
\(\mathrm{~m}^2\lt 1\)
\(|\mathrm{~m}|\lt 1\)
AP EAMCET-21.09.2020
Ellipse
120570
\(S\) and \(T\) are the foci of an ellipse and \(B\) is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Equation of the ellipse is -
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Foci are \(\mathrm{S}(\mathrm{ae}, 0) \mathrm{T}(-\mathrm{ae}, 0)\) \(\mathrm{B}(0, b)\) is the end of the minor axis.
\(\mathrm{SB}=\mathrm{ST}\)
\(\mathrm{SB}^2=\mathrm{ST}^2\)
\(\mathrm{~b}^2=3 \mathrm{a}^2 \mathrm{e}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=3 \mathrm{a}^2 \mathrm{e}^2\)
\(1-\mathrm{e}^2=3 \mathrm{e}^2\)
\(4 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{4}\)
\(\mathrm{e}=\frac{1}{2}\)Eccentricity of the ellipse, \(\mathrm{e}=\frac{1}{2}\)
WB JEE-2019
Ellipse
120571
If \(B\) and \(B\) ' are the ends of minor axis and \(S\) and \(S^{\prime}\) are the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) then the area of the rhombus SBS'B' will be
1 12 sq units
2 48 sq units
3 24 sq units
4 36 sq units
Explanation:
C Given,
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{25}}\)
\(\mathrm{e}=\frac{4}{5}\)
\(\therefore \mathrm{S}=(\mathrm{ae}, 0)=\left(5 \times \frac{4}{5}, 0\right)\)
\(\mathrm{S}=(4,0)\)
Area of rhombus \(=4 \times\) area of \(\Delta \mathrm{BOS}\),
\(=4 \times\left(\frac{1}{2} \times \mathrm{OS} \times \mathrm{OB}\right)\)
\(=4 \times \frac{1}{2} \times 4 \times 3\)
\(\mathrm{~A} =24 \text { sq. unit }\)
WB JEE-2020
Ellipse
120572
The equation \(\operatorname{rcos}\left(\theta-\frac{\pi}{3}\right)=2\) represents
1 a circle
2 a parabola
3 an ellipse
4 a straight line
Explanation:
D Given,
\(r \cos \left(\theta-\frac{\pi}{3}\right)=2\)
\(r\left[\cos \theta \cos \frac{\pi}{3}+\sin \theta \sin \frac{\pi}{3}\right]=2\)
\(\mathrm{r} \cos \theta \cdot \frac{1}{2}+\mathrm{r} \sin \theta \cdot \frac{\sqrt{3}}{2}=2\)
Let, \(\mathrm{r} \cos \theta=\mathrm{x}, \mathrm{y}=\mathrm{r} \sin \theta\)
Therefore, \(x+\sqrt{3} y=4\)
The equation is represents a straight line
120568
An ellipse having the coordinate axes as its axes and its major axes along \(Y\)-axis, passes through the point \((-3,1)\) and has eccentricity \(\sqrt{\frac{2}{5}}\). Then its equation is
1 \(3 x^2+5 y^2-15=0\)
2 \(5 x^2+3 y^2-32=0\)
3 \(3 x^2+5 y^2-32=0\)
4 \(5 \mathrm{x}^2+3 \mathrm{y}^2-48=0\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{\frac{2}{5}}\)
\(\sqrt{1-\frac{b^2}{a^2}}=\sqrt{\frac{2}{5}}\)
\(5\left(a^2-b^2\right)=2 a^2\)
\(a^2=\frac{5}{3} b^2\)
Equation of ellipse with point \((-3,1)\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1\)
\(\frac{9}{a^2}+\frac{1}{b^2}=1\)
\(9 b^2+a^2=a^2 b^2\)
\(9 b^2+\frac{5 b^2}{3}=\frac{5}{3} b^4 \quad \text { \{from (i)\} }\)
\(\frac{27 b^2+5 b^2}{3}=\frac{5}{3} b^4\)
\(\frac{5}{3} b^2=\frac{32}{3}\)
\(b^2=\frac{32}{5} \text { and } a^2=\frac{32}{3}\)
\{from (i) \(\}\)
Therefore, the equation of ellipse -
\(\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1\)
\(\frac{3 x^2}{32}+\frac{5 y^2}{32}=1\)
\(3 x^2+5 y^2-32=0\)
AP EAMCET-22.04.2018
Ellipse
120569
The line \(x=m\) meets an ellipse \(9 x^2+y^2=9\) in the real and distinct points if and only if
1 \(|\mathrm{m}|>1\)
2 \(|\mathrm{m}|\lt 1\)
3 \(|\mathrm{m}|>2\)
4 \(|\mathrm{m}|\lt 2\)
Explanation:
B Given,
\(9 x^2+y^2=9\)
\(\frac{x^2}{1}+\frac{y^2}{9}=1\)
Also, \(\mathrm{x}=\mathrm{m} \quad\) (put in \(\left.\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\)
So, \(\quad \mathrm{m}^2+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{y}^2=9\left(1-\mathrm{m}^2\right)\)
It gives real and distinct if,
\(1-\mathrm{m}^2>0\)
\(\mathrm{~m}^2\lt 1\)
\(|\mathrm{~m}|\lt 1\)
AP EAMCET-21.09.2020
Ellipse
120570
\(S\) and \(T\) are the foci of an ellipse and \(B\) is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Equation of the ellipse is -
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Foci are \(\mathrm{S}(\mathrm{ae}, 0) \mathrm{T}(-\mathrm{ae}, 0)\) \(\mathrm{B}(0, b)\) is the end of the minor axis.
\(\mathrm{SB}=\mathrm{ST}\)
\(\mathrm{SB}^2=\mathrm{ST}^2\)
\(\mathrm{~b}^2=3 \mathrm{a}^2 \mathrm{e}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=3 \mathrm{a}^2 \mathrm{e}^2\)
\(1-\mathrm{e}^2=3 \mathrm{e}^2\)
\(4 \mathrm{e}^2=1\)
\(\mathrm{e}^2=\frac{1}{4}\)
\(\mathrm{e}=\frac{1}{2}\)Eccentricity of the ellipse, \(\mathrm{e}=\frac{1}{2}\)
WB JEE-2019
Ellipse
120571
If \(B\) and \(B\) ' are the ends of minor axis and \(S\) and \(S^{\prime}\) are the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) then the area of the rhombus SBS'B' will be
1 12 sq units
2 48 sq units
3 24 sq units
4 36 sq units
Explanation:
C Given,
\(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{25}}\)
\(\mathrm{e}=\frac{4}{5}\)
\(\therefore \mathrm{S}=(\mathrm{ae}, 0)=\left(5 \times \frac{4}{5}, 0\right)\)
\(\mathrm{S}=(4,0)\)
Area of rhombus \(=4 \times\) area of \(\Delta \mathrm{BOS}\),
\(=4 \times\left(\frac{1}{2} \times \mathrm{OS} \times \mathrm{OB}\right)\)
\(=4 \times \frac{1}{2} \times 4 \times 3\)
\(\mathrm{~A} =24 \text { sq. unit }\)
WB JEE-2020
Ellipse
120572
The equation \(\operatorname{rcos}\left(\theta-\frac{\pi}{3}\right)=2\) represents
1 a circle
2 a parabola
3 an ellipse
4 a straight line
Explanation:
D Given,
\(r \cos \left(\theta-\frac{\pi}{3}\right)=2\)
\(r\left[\cos \theta \cos \frac{\pi}{3}+\sin \theta \sin \frac{\pi}{3}\right]=2\)
\(\mathrm{r} \cos \theta \cdot \frac{1}{2}+\mathrm{r} \sin \theta \cdot \frac{\sqrt{3}}{2}=2\)
Let, \(\mathrm{r} \cos \theta=\mathrm{x}, \mathrm{y}=\mathrm{r} \sin \theta\)
Therefore, \(x+\sqrt{3} y=4\)
The equation is represents a straight line
