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Ellipse

120573 The equation of the ellipse having a vertex at $(6, 1)$ a focus at $(4, 1)$ and the eccentricity $\frac{3}{5}$ is

1

\frac{(x - 1)^{2}}{16} + \frac{(y - 1)^{2}}{25} = 1

2

\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1

3

\frac{(x + 1)^{2}}{25} + \frac{(y + 1)^{2}}{16} = 1

4

\frac{(x + 1)^{2}}{16} + \frac{(y + 1)^{2}}{25} = 1

Explanation:

B Given, vertex $(6, 1)$ and focus $(4, 1)$
$e = \frac{3}{5}, SA = a - ae$
$2 = a (1 - e)$
Also, $2 = a (1 - e)$
$2 = a (1 - \frac{3}{5})$
$a = 5$
$∴ b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{9}{25}) = 16$ Hence, the required equation $\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1$

AP EAMCET-23.04.2018

Ellipse

120574 Find the equation of an ellipse whose vertices are $(\pm 5, 0)$ and foci are $(\pm 4, 0)$

1

9 x^{2} + 25 y^{2} = 225

2

25 x^{2} + 9 y^{2} = 225

3

3 x^{2} + 4 y^{2} = 192

4

4 x^{2} + 3 y^{2} = 12

Explanation:

A Given, vertices $(\pm 5, 0) = (\pm a, 0)$
$Foci (\pm 4, 0) = (\pm ae, 0)$
$ae = 4$
$e = \frac{4}{a} = \frac{4}{5}$
$∴ b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{16}{25}) = 9$
$b = 3$
Hence, equation: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
$9 x^{2} + 25 y^{2} = 225$

AP EAMCET-05.10.2021

Ellipse

120575 If $α, β$ are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^{2} + 4 y^{2} - 4$ then $\sqrt{3} \cos \frac{α + β}{2} =$

1

2 \cos \frac{α - β}{2}

2

2 \sin \frac{α - β}{2}

3

2 \sec \frac{α + β}{2}

4

2 \sin \frac{α + β}{2}

Explanation:

A Given,
$x^{2} + 4 y^{2} - 4 = 0$
$\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$
Eccentricity, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1}{4}}$
$e = \frac{\sqrt{3}}{2}$
$∴ Focus = (\sqrt{3}, 0)$
Equation of required chord.
$\frac{x}{a} \cos (\frac{α + β}{2}) + \frac{y}{b} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
$\frac{x}{2} \cos (\frac{α + β}{2}) + \frac{y}{1} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
It passes through $(\sqrt{3}, 0)$
$\frac{\sqrt{3}}{2} \cos (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
$\sqrt{3} \cos (\frac{α + β}{2}) = 2 \cos (\frac{α - β}{2})$

AP EAMCET-24.04.2018

Ellipse

120576 In the ellipse, if the distance between the foci is 6 units and the length of its minor axis is 8 units, then its eccentricity is

1

\frac{1}{2}

2

\frac{7}{5}

3

\frac{1}{\sqrt{5}}

4

\frac{3}{5}

Explanation:

D Given, $2 ae = 6$
And, minor axis
$2 b = 8$
We known that,
$b^{2} = a^{2} (1 - e^{2})$
$(4)^{2} = a^{2} - a^{2} e^{2}$
$(4)^{2} = a^{2} - (3)^{2}$
$a^{2} = 25$
$a = 5$
$2 a e = 6$
$e = \frac{3}{5}$
and
$∴ e = \frac{3}{5}$

AP EAMCET-19.08.2021

Ellipse

120573 The equation of the ellipse having a vertex at $(6, 1)$ a focus at $(4, 1)$ and the eccentricity $\frac{3}{5}$ is

1

\frac{(x - 1)^{2}}{16} + \frac{(y - 1)^{2}}{25} = 1

2

\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1

3

\frac{(x + 1)^{2}}{25} + \frac{(y + 1)^{2}}{16} = 1

4

\frac{(x + 1)^{2}}{16} + \frac{(y + 1)^{2}}{25} = 1

Explanation:

B Given, vertex $(6, 1)$ and focus $(4, 1)$
$e = \frac{3}{5}, SA = a - ae$
$2 = a (1 - e)$
Also, $2 = a (1 - e)$
$2 = a (1 - \frac{3}{5})$
$a = 5$
$∴ b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{9}{25}) = 16$ Hence, the required equation $\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1$

AP EAMCET-23.04.2018

Ellipse

120574 Find the equation of an ellipse whose vertices are $(\pm 5, 0)$ and foci are $(\pm 4, 0)$

1

9 x^{2} + 25 y^{2} = 225

2

25 x^{2} + 9 y^{2} = 225

3

3 x^{2} + 4 y^{2} = 192

4

4 x^{2} + 3 y^{2} = 12

Explanation:

A Given, vertices $(\pm 5, 0) = (\pm a, 0)$
$Foci (\pm 4, 0) = (\pm ae, 0)$
$ae = 4$
$e = \frac{4}{a} = \frac{4}{5}$
$∴ b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{16}{25}) = 9$
$b = 3$
Hence, equation: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
$9 x^{2} + 25 y^{2} = 225$

AP EAMCET-05.10.2021

Ellipse

120575 If $α, β$ are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^{2} + 4 y^{2} - 4$ then $\sqrt{3} \cos \frac{α + β}{2} =$

1

2 \cos \frac{α - β}{2}

2

2 \sin \frac{α - β}{2}

3

2 \sec \frac{α + β}{2}

4

2 \sin \frac{α + β}{2}

Explanation:

A Given,
$x^{2} + 4 y^{2} - 4 = 0$
$\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$
Eccentricity, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1}{4}}$
$e = \frac{\sqrt{3}}{2}$
$∴ Focus = (\sqrt{3}, 0)$
Equation of required chord.
$\frac{x}{a} \cos (\frac{α + β}{2}) + \frac{y}{b} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
$\frac{x}{2} \cos (\frac{α + β}{2}) + \frac{y}{1} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
It passes through $(\sqrt{3}, 0)$
$\frac{\sqrt{3}}{2} \cos (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
$\sqrt{3} \cos (\frac{α + β}{2}) = 2 \cos (\frac{α - β}{2})$

AP EAMCET-24.04.2018

Ellipse

120576 In the ellipse, if the distance between the foci is 6 units and the length of its minor axis is 8 units, then its eccentricity is

1

\frac{1}{2}

2

\frac{7}{5}

3

\frac{1}{\sqrt{5}}

4

\frac{3}{5}

Explanation:

D Given, $2 ae = 6$
And, minor axis
$2 b = 8$
We known that,
$b^{2} = a^{2} (1 - e^{2})$
$(4)^{2} = a^{2} - a^{2} e^{2}$
$(4)^{2} = a^{2} - (3)^{2}$
$a^{2} = 25$
$a = 5$
$2 a e = 6$
$e = \frac{3}{5}$
and
$∴ e = \frac{3}{5}$

AP EAMCET-19.08.2021

Ellipse

120573 The equation of the ellipse having a vertex at $(6, 1)$ a focus at $(4, 1)$ and the eccentricity $\frac{3}{5}$ is

1

\frac{(x - 1)^{2}}{16} + \frac{(y - 1)^{2}}{25} = 1

2

\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1

3

\frac{(x + 1)^{2}}{25} + \frac{(y + 1)^{2}}{16} = 1

4

\frac{(x + 1)^{2}}{16} + \frac{(y + 1)^{2}}{25} = 1

Explanation:

B Given, vertex $(6, 1)$ and focus $(4, 1)$
$e = \frac{3}{5}, SA = a - ae$
$2 = a (1 - e)$
Also, $2 = a (1 - e)$
$2 = a (1 - \frac{3}{5})$
$a = 5$
$∴ b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{9}{25}) = 16$ Hence, the required equation $\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1$

AP EAMCET-23.04.2018

Ellipse

120574 Find the equation of an ellipse whose vertices are $(\pm 5, 0)$ and foci are $(\pm 4, 0)$

1

9 x^{2} + 25 y^{2} = 225

2

25 x^{2} + 9 y^{2} = 225

3

3 x^{2} + 4 y^{2} = 192

4

4 x^{2} + 3 y^{2} = 12

Explanation:

A Given, vertices $(\pm 5, 0) = (\pm a, 0)$
$Foci (\pm 4, 0) = (\pm ae, 0)$
$ae = 4$
$e = \frac{4}{a} = \frac{4}{5}$
$∴ b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{16}{25}) = 9$
$b = 3$
Hence, equation: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
$9 x^{2} + 25 y^{2} = 225$

AP EAMCET-05.10.2021

Ellipse

120575 If $α, β$ are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^{2} + 4 y^{2} - 4$ then $\sqrt{3} \cos \frac{α + β}{2} =$

1

2 \cos \frac{α - β}{2}

2

2 \sin \frac{α - β}{2}

3

2 \sec \frac{α + β}{2}

4

2 \sin \frac{α + β}{2}

Explanation:

A Given,
$x^{2} + 4 y^{2} - 4 = 0$
$\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$
Eccentricity, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1}{4}}$
$e = \frac{\sqrt{3}}{2}$
$∴ Focus = (\sqrt{3}, 0)$
Equation of required chord.
$\frac{x}{a} \cos (\frac{α + β}{2}) + \frac{y}{b} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
$\frac{x}{2} \cos (\frac{α + β}{2}) + \frac{y}{1} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
It passes through $(\sqrt{3}, 0)$
$\frac{\sqrt{3}}{2} \cos (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
$\sqrt{3} \cos (\frac{α + β}{2}) = 2 \cos (\frac{α - β}{2})$

AP EAMCET-24.04.2018

Ellipse

120576 In the ellipse, if the distance between the foci is 6 units and the length of its minor axis is 8 units, then its eccentricity is

1

\frac{1}{2}

2

\frac{7}{5}

3

\frac{1}{\sqrt{5}}

4

\frac{3}{5}

Explanation:

D Given, $2 ae = 6$
And, minor axis
$2 b = 8$
We known that,
$b^{2} = a^{2} (1 - e^{2})$
$(4)^{2} = a^{2} - a^{2} e^{2}$
$(4)^{2} = a^{2} - (3)^{2}$
$a^{2} = 25$
$a = 5$
$2 a e = 6$
$e = \frac{3}{5}$
and
$∴ e = \frac{3}{5}$

AP EAMCET-19.08.2021

Ellipse

120573 The equation of the ellipse having a vertex at $(6, 1)$ a focus at $(4, 1)$ and the eccentricity $\frac{3}{5}$ is

1

\frac{(x - 1)^{2}}{16} + \frac{(y - 1)^{2}}{25} = 1

2

\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1

3

\frac{(x + 1)^{2}}{25} + \frac{(y + 1)^{2}}{16} = 1

4

\frac{(x + 1)^{2}}{16} + \frac{(y + 1)^{2}}{25} = 1

Explanation:

B Given, vertex $(6, 1)$ and focus $(4, 1)$
$e = \frac{3}{5}, SA = a - ae$
$2 = a (1 - e)$
Also, $2 = a (1 - e)$
$2 = a (1 - \frac{3}{5})$
$a = 5$
$∴ b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{9}{25}) = 16$ Hence, the required equation $\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1$

AP EAMCET-23.04.2018

Ellipse

120574 Find the equation of an ellipse whose vertices are $(\pm 5, 0)$ and foci are $(\pm 4, 0)$

1

9 x^{2} + 25 y^{2} = 225

2

25 x^{2} + 9 y^{2} = 225

3

3 x^{2} + 4 y^{2} = 192

4

4 x^{2} + 3 y^{2} = 12

Explanation:

A Given, vertices $(\pm 5, 0) = (\pm a, 0)$
$Foci (\pm 4, 0) = (\pm ae, 0)$
$ae = 4$
$e = \frac{4}{a} = \frac{4}{5}$
$∴ b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{16}{25}) = 9$
$b = 3$
Hence, equation: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
$9 x^{2} + 25 y^{2} = 225$

AP EAMCET-05.10.2021

Ellipse

120575 If $α, β$ are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^{2} + 4 y^{2} - 4$ then $\sqrt{3} \cos \frac{α + β}{2} =$

1

2 \cos \frac{α - β}{2}

2

2 \sin \frac{α - β}{2}

3

2 \sec \frac{α + β}{2}

4

2 \sin \frac{α + β}{2}

Explanation:

A Given,
$x^{2} + 4 y^{2} - 4 = 0$
$\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$
Eccentricity, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1}{4}}$
$e = \frac{\sqrt{3}}{2}$
$∴ Focus = (\sqrt{3}, 0)$
Equation of required chord.
$\frac{x}{a} \cos (\frac{α + β}{2}) + \frac{y}{b} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
$\frac{x}{2} \cos (\frac{α + β}{2}) + \frac{y}{1} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
It passes through $(\sqrt{3}, 0)$
$\frac{\sqrt{3}}{2} \cos (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$
$\sqrt{3} \cos (\frac{α + β}{2}) = 2 \cos (\frac{α - β}{2})$

AP EAMCET-24.04.2018

Ellipse

120576 In the ellipse, if the distance between the foci is 6 units and the length of its minor axis is 8 units, then its eccentricity is

1

\frac{1}{2}

2

\frac{7}{5}

3

\frac{1}{\sqrt{5}}

4

\frac{3}{5}

Explanation:

D Given, $2 ae = 6$
And, minor axis
$2 b = 8$
We known that,
$b^{2} = a^{2} (1 - e^{2})$
$(4)^{2} = a^{2} - a^{2} e^{2}$
$(4)^{2} = a^{2} - (3)^{2}$
$a^{2} = 25$
$a = 5$
$2 a e = 6$
$e = \frac{3}{5}$
and
$∴ e = \frac{3}{5}$

AP EAMCET-19.08.2021

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