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Ellipse

120577 If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is

1

\frac{1}{4} (\sqrt{5} - 1)

2

\frac{1}{2} (\sqrt{5} + 1)

3

\frac{1}{2} (\sqrt{5} - 1)

4

\frac{1}{4} (\sqrt{5} + 1)

Explanation:

C According to question,
The difference between foci = length of latus rectum
$2 a e = \frac{2 b^{2}}{a}$
$e = \frac{b^{2}}{a^{2}}$
Also, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 - e}$
$e^{2} = 1 - e$
$e^{2} + e - 1 = 0$
And
$0 < e < 1$
$e = \frac{- 1 \pm \sqrt{1 + 4}}{2}$
$e = \frac{1}{2} (\sqrt{5} - 1)$
(only positive value)

WB JEE-2013

Ellipse

120578 The equation of the ellipse whose foci are $(\pm, 2, 0)$ and eccentricity $\frac{1}{2}$ is:

1

\frac{x^{2}}{12} + \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1

3

\frac{x^{2}}{16} + \frac{y^{2}}{8} = 1

4 None of these

Explanation:

B Given, $e = \frac{1}{2}$
And foci $(\pm ae, 0) = (\pm 2, 0)$
$∴ a \times \frac{1}{2} = 2$
$∴ a = 4$
As we know that,
$e^{2} = \frac{a^{2} - b^{2}}{a^{2}}$
$\frac{1}{4} = \frac{16 - b^{2}}{16}$
$b^{2} = 12$
So, the equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$

Jamia Millia Islamia-2005

Ellipse

120579 $x = 4 (1 + \cos θ)$ and $y = 3 (1 + \sin θ)$ are the parametric equation of

1

\frac{(x - 3)^{2}}{9} + \frac{(y - 4)^{2}}{16} = 1

2

\frac{(x + 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1

3

\frac{(x - 4)^{2}}{16} - \frac{(y - 3)^{2}}{9} = 1

4

\frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1

Explanation:

D Given,
$x = 4 (1 + \cos θ)$
$y = 3 (1 + \sin θ)$
From eqn. (i) \& (ii), we get-
$\cos θ = \frac{x}{4} - 1$
$\sin θ = \frac{y}{3} - 1$
Squaring and adding equation (iii) \& (iv)
$\cos^{2} θ + \sin^{2} θ = {[\frac{x}{4} - 1]}^{2} + {[\frac{y}{3} - 1]}^{2}$
$1 = \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9}$
$∴ \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1$

Jamia Millia Islamia-2008

Ellipse

120580 The sum of major and minor axes lengths of an ellipse whose eccentricity is $4 / 5$ and length of latus rectum is 14.4 is

1 24

2 32

3 64

4 48

Explanation:

C Given,
Length of major axis of an ellipse $= 2 a$
Length of minor axis of an ellipse $= 2 b$
Eccentricity $(e) = \frac{4}{5}$
And length of latus rectum $= 14.4$
$\frac{2 b^{2}}{a} = 14.4$
$b^{2} = 7.2 a$
$Also, b^{2} = a^{2} (1 - e^{2})$
$7 \cdot 2 a = a^{2} [1 - {(\frac{4}{5})}^{2}]$
$7.2 a = a^{2} (\frac{9}{25})$
$a = 0 or a = 20$
$But a \neq 0 a = 20$
$Putting a = 20 in b^{2} = 7.2 a, we get-$
$b = 12$
The sum of major and minor axis $= 2 a + 2 b$
$= 2 (20 + 12)$
$= 2 (32)$
$= 64 units.$
But $a \neq 0 a = 20$

J and K CET-2015

Ellipse

120577 If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is

1

\frac{1}{4} (\sqrt{5} - 1)

2

\frac{1}{2} (\sqrt{5} + 1)

3

\frac{1}{2} (\sqrt{5} - 1)

4

\frac{1}{4} (\sqrt{5} + 1)

Explanation:

C According to question,
The difference between foci = length of latus rectum
$2 a e = \frac{2 b^{2}}{a}$
$e = \frac{b^{2}}{a^{2}}$
Also, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 - e}$
$e^{2} = 1 - e$
$e^{2} + e - 1 = 0$
And
$0 < e < 1$
$e = \frac{- 1 \pm \sqrt{1 + 4}}{2}$
$e = \frac{1}{2} (\sqrt{5} - 1)$
(only positive value)

WB JEE-2013

Ellipse

120578 The equation of the ellipse whose foci are $(\pm, 2, 0)$ and eccentricity $\frac{1}{2}$ is:

1

\frac{x^{2}}{12} + \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1

3

\frac{x^{2}}{16} + \frac{y^{2}}{8} = 1

4 None of these

Explanation:

B Given, $e = \frac{1}{2}$
And foci $(\pm ae, 0) = (\pm 2, 0)$
$∴ a \times \frac{1}{2} = 2$
$∴ a = 4$
As we know that,
$e^{2} = \frac{a^{2} - b^{2}}{a^{2}}$
$\frac{1}{4} = \frac{16 - b^{2}}{16}$
$b^{2} = 12$
So, the equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$

Jamia Millia Islamia-2005

Ellipse

120579 $x = 4 (1 + \cos θ)$ and $y = 3 (1 + \sin θ)$ are the parametric equation of

1

\frac{(x - 3)^{2}}{9} + \frac{(y - 4)^{2}}{16} = 1

2

\frac{(x + 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1

3

\frac{(x - 4)^{2}}{16} - \frac{(y - 3)^{2}}{9} = 1

4

\frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1

Explanation:

D Given,
$x = 4 (1 + \cos θ)$
$y = 3 (1 + \sin θ)$
From eqn. (i) \& (ii), we get-
$\cos θ = \frac{x}{4} - 1$
$\sin θ = \frac{y}{3} - 1$
Squaring and adding equation (iii) \& (iv)
$\cos^{2} θ + \sin^{2} θ = {[\frac{x}{4} - 1]}^{2} + {[\frac{y}{3} - 1]}^{2}$
$1 = \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9}$
$∴ \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1$

Jamia Millia Islamia-2008

Ellipse

120580 The sum of major and minor axes lengths of an ellipse whose eccentricity is $4 / 5$ and length of latus rectum is 14.4 is

1 24

2 32

3 64

4 48

Explanation:

C Given,
Length of major axis of an ellipse $= 2 a$
Length of minor axis of an ellipse $= 2 b$
Eccentricity $(e) = \frac{4}{5}$
And length of latus rectum $= 14.4$
$\frac{2 b^{2}}{a} = 14.4$
$b^{2} = 7.2 a$
$Also, b^{2} = a^{2} (1 - e^{2})$
$7 \cdot 2 a = a^{2} [1 - {(\frac{4}{5})}^{2}]$
$7.2 a = a^{2} (\frac{9}{25})$
$a = 0 or a = 20$
$But a \neq 0 a = 20$
$Putting a = 20 in b^{2} = 7.2 a, we get-$
$b = 12$
The sum of major and minor axis $= 2 a + 2 b$
$= 2 (20 + 12)$
$= 2 (32)$
$= 64 units.$
But $a \neq 0 a = 20$

J and K CET-2015

Ellipse

120577 If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is

1

\frac{1}{4} (\sqrt{5} - 1)

2

\frac{1}{2} (\sqrt{5} + 1)

3

\frac{1}{2} (\sqrt{5} - 1)

4

\frac{1}{4} (\sqrt{5} + 1)

Explanation:

C According to question,
The difference between foci = length of latus rectum
$2 a e = \frac{2 b^{2}}{a}$
$e = \frac{b^{2}}{a^{2}}$
Also, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 - e}$
$e^{2} = 1 - e$
$e^{2} + e - 1 = 0$
And
$0 < e < 1$
$e = \frac{- 1 \pm \sqrt{1 + 4}}{2}$
$e = \frac{1}{2} (\sqrt{5} - 1)$
(only positive value)

WB JEE-2013

Ellipse

120578 The equation of the ellipse whose foci are $(\pm, 2, 0)$ and eccentricity $\frac{1}{2}$ is:

1

\frac{x^{2}}{12} + \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1

3

\frac{x^{2}}{16} + \frac{y^{2}}{8} = 1

4 None of these

Explanation:

B Given, $e = \frac{1}{2}$
And foci $(\pm ae, 0) = (\pm 2, 0)$
$∴ a \times \frac{1}{2} = 2$
$∴ a = 4$
As we know that,
$e^{2} = \frac{a^{2} - b^{2}}{a^{2}}$
$\frac{1}{4} = \frac{16 - b^{2}}{16}$
$b^{2} = 12$
So, the equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$

Jamia Millia Islamia-2005

Ellipse

120579 $x = 4 (1 + \cos θ)$ and $y = 3 (1 + \sin θ)$ are the parametric equation of

1

\frac{(x - 3)^{2}}{9} + \frac{(y - 4)^{2}}{16} = 1

2

\frac{(x + 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1

3

\frac{(x - 4)^{2}}{16} - \frac{(y - 3)^{2}}{9} = 1

4

\frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1

Explanation:

D Given,
$x = 4 (1 + \cos θ)$
$y = 3 (1 + \sin θ)$
From eqn. (i) \& (ii), we get-
$\cos θ = \frac{x}{4} - 1$
$\sin θ = \frac{y}{3} - 1$
Squaring and adding equation (iii) \& (iv)
$\cos^{2} θ + \sin^{2} θ = {[\frac{x}{4} - 1]}^{2} + {[\frac{y}{3} - 1]}^{2}$
$1 = \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9}$
$∴ \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1$

Jamia Millia Islamia-2008

Ellipse

120580 The sum of major and minor axes lengths of an ellipse whose eccentricity is $4 / 5$ and length of latus rectum is 14.4 is

1 24

2 32

3 64

4 48

Explanation:

C Given,
Length of major axis of an ellipse $= 2 a$
Length of minor axis of an ellipse $= 2 b$
Eccentricity $(e) = \frac{4}{5}$
And length of latus rectum $= 14.4$
$\frac{2 b^{2}}{a} = 14.4$
$b^{2} = 7.2 a$
$Also, b^{2} = a^{2} (1 - e^{2})$
$7 \cdot 2 a = a^{2} [1 - {(\frac{4}{5})}^{2}]$
$7.2 a = a^{2} (\frac{9}{25})$
$a = 0 or a = 20$
$But a \neq 0 a = 20$
$Putting a = 20 in b^{2} = 7.2 a, we get-$
$b = 12$
The sum of major and minor axis $= 2 a + 2 b$
$= 2 (20 + 12)$
$= 2 (32)$
$= 64 units.$
But $a \neq 0 a = 20$

J and K CET-2015

Ellipse

120577 If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is

1

\frac{1}{4} (\sqrt{5} - 1)

2

\frac{1}{2} (\sqrt{5} + 1)

3

\frac{1}{2} (\sqrt{5} - 1)

4

\frac{1}{4} (\sqrt{5} + 1)

Explanation:

C According to question,
The difference between foci = length of latus rectum
$2 a e = \frac{2 b^{2}}{a}$
$e = \frac{b^{2}}{a^{2}}$
Also, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 - e}$
$e^{2} = 1 - e$
$e^{2} + e - 1 = 0$
And
$0 < e < 1$
$e = \frac{- 1 \pm \sqrt{1 + 4}}{2}$
$e = \frac{1}{2} (\sqrt{5} - 1)$
(only positive value)

WB JEE-2013

Ellipse

120578 The equation of the ellipse whose foci are $(\pm, 2, 0)$ and eccentricity $\frac{1}{2}$ is:

1

\frac{x^{2}}{12} + \frac{y^{2}}{16} = 1

2

\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1

3

\frac{x^{2}}{16} + \frac{y^{2}}{8} = 1

4 None of these

Explanation:

B Given, $e = \frac{1}{2}$
And foci $(\pm ae, 0) = (\pm 2, 0)$
$∴ a \times \frac{1}{2} = 2$
$∴ a = 4$
As we know that,
$e^{2} = \frac{a^{2} - b^{2}}{a^{2}}$
$\frac{1}{4} = \frac{16 - b^{2}}{16}$
$b^{2} = 12$
So, the equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$

Jamia Millia Islamia-2005

Ellipse

120579 $x = 4 (1 + \cos θ)$ and $y = 3 (1 + \sin θ)$ are the parametric equation of

1

\frac{(x - 3)^{2}}{9} + \frac{(y - 4)^{2}}{16} = 1

2

\frac{(x + 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1

3

\frac{(x - 4)^{2}}{16} - \frac{(y - 3)^{2}}{9} = 1

4

\frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1

Explanation:

D Given,
$x = 4 (1 + \cos θ)$
$y = 3 (1 + \sin θ)$
From eqn. (i) \& (ii), we get-
$\cos θ = \frac{x}{4} - 1$
$\sin θ = \frac{y}{3} - 1$
Squaring and adding equation (iii) \& (iv)
$\cos^{2} θ + \sin^{2} θ = {[\frac{x}{4} - 1]}^{2} + {[\frac{y}{3} - 1]}^{2}$
$1 = \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9}$
$∴ \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1$

Jamia Millia Islamia-2008

Ellipse

120580 The sum of major and minor axes lengths of an ellipse whose eccentricity is $4 / 5$ and length of latus rectum is 14.4 is

1 24

2 32

3 64

4 48

Explanation:

C Given,
Length of major axis of an ellipse $= 2 a$
Length of minor axis of an ellipse $= 2 b$
Eccentricity $(e) = \frac{4}{5}$
And length of latus rectum $= 14.4$
$\frac{2 b^{2}}{a} = 14.4$
$b^{2} = 7.2 a$
$Also, b^{2} = a^{2} (1 - e^{2})$
$7 \cdot 2 a = a^{2} [1 - {(\frac{4}{5})}^{2}]$
$7.2 a = a^{2} (\frac{9}{25})$
$a = 0 or a = 20$
$But a \neq 0 a = 20$
$Putting a = 20 in b^{2} = 7.2 a, we get-$
$b = 12$
The sum of major and minor axis $= 2 a + 2 b$
$= 2 (20 + 12)$
$= 2 (32)$
$= 64 units.$
But $a \neq 0 a = 20$

J and K CET-2015