QBManager

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
$∴ 2 ae = 2$
$2 a = 10$
$a = 5$
$a^{2} = 25$
$∴ 2 \times 5 \times e = 2 e = \frac{1}{5}$
$∵ b^{2} = a^{2} (1 - e^{2}) b^{2} = 25 (1 - \frac{1}{25})$
$b^{2} = 24$
we know that, centre of ellipse $=$ mid point of foci
$= (\frac{2 + 4}{2}, \frac{2 + 2}{2}) = (3, 2)$
Hence, equation of ellipse is
$\frac{(x - 3)^{2}}{25} + \frac{(y - 2)^{2}}{24} = 1$

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 $(m_{1} \times m_{2}) = \frac{- b^{2}}{a^{2}}$
Where $m_{1}$ and $m_{2}$ are slopes of the respective diameters
$1 \times (\frac{- 2}{3}) = \frac{- b^{2}}{a^{2}}$
$b^{2} = \frac{2}{3} a^{2}$
How, $b^{2} = a^{2} (1 - e^{2}) = \frac{2}{3} a^{2} = a^{2} (1 - e^{2})$
$\frac{2}{3} = 1 - e^{2}$
$e^{2} = \frac{1}{3} = \frac{1}{\sqrt{3}}$

JCECE-2012

Ellipse

120519 The length of latusrectum of the ellipse $2 x^{2} + y^{2}$ $- 8 x + 2 y + 7 = 0$ , is

1

\sqrt{2}

2 2

3 8

4 None of these

Explanation:

A Given,
$2 x^{2} + y^{2} - 8 x + 2 y + 7 = 0$
$2 x^{2} - 8 x + y^{2} + 2 y + 7 = 0$
$2 (x^{2} - 4 x + 4) - 8 + (y^{2} + 2 y + 1) - 1 + 7 = 0$
$2 (x - 2)^{2} + (y + 1)^{2} = 2$
$\frac{(x - 2)^{2}}{1} + \frac{(y + 1)^{2}}{2} = 1$ Hence, Length of latusrectum $= \frac{2 a^{2}}{b} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$

JCECE-2010

Ellipse

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
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$∴$ The general equation of the tangent of the ellipse is
$y = mx \pm \sqrt{a^{2} m^{2} + b^{2}}$
Now,
$3 x - 2 y - 20 = 0$
$∴ y = \frac{3}{2} x - 10$ is a tangent to the ellipse
Therefore, on comparing with equation (i), we get-
$m = \frac{3}{2} and a^{2} m^{2} + b^{2} = 100$
$a^{2} \cdot \frac{9}{4} + b^{2} = 100$
$9 a^{2} + 4 b^{2} = 400$
Similarly, line $x + 6 y - 20 = 0$
$y = - \frac{1}{6} x + \frac{10}{3}$
is a tangent to the ellipse is therefore on comparing with equation(i), we get :-
$m = - \frac{1}{6} and a^{2} m^{2} + b^{2} = \frac{100}{9}$
$\frac{a^{2}}{36} + b^{2} = \frac{100}{9}$
$a^{2} + 36 b^{2} = 400$
From equation (ii) and (iii), we get-
$a^{2} = 40 and b^{2} = 10$ Hence, $\frac{x^{2}}{40} + \frac{y^{2}}{10} = 1$ is required equations of ellipse.

JCECE-2010

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
$∴ 2 ae = 2$
$2 a = 10$
$a = 5$
$a^{2} = 25$
$∴ 2 \times 5 \times e = 2 e = \frac{1}{5}$
$∵ b^{2} = a^{2} (1 - e^{2}) b^{2} = 25 (1 - \frac{1}{25})$
$b^{2} = 24$
we know that, centre of ellipse $=$ mid point of foci
$= (\frac{2 + 4}{2}, \frac{2 + 2}{2}) = (3, 2)$
Hence, equation of ellipse is
$\frac{(x - 3)^{2}}{25} + \frac{(y - 2)^{2}}{24} = 1$

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 $(m_{1} \times m_{2}) = \frac{- b^{2}}{a^{2}}$
Where $m_{1}$ and $m_{2}$ are slopes of the respective diameters
$1 \times (\frac{- 2}{3}) = \frac{- b^{2}}{a^{2}}$
$b^{2} = \frac{2}{3} a^{2}$
How, $b^{2} = a^{2} (1 - e^{2}) = \frac{2}{3} a^{2} = a^{2} (1 - e^{2})$
$\frac{2}{3} = 1 - e^{2}$
$e^{2} = \frac{1}{3} = \frac{1}{\sqrt{3}}$

JCECE-2012

Ellipse

120519 The length of latusrectum of the ellipse $2 x^{2} + y^{2}$ $- 8 x + 2 y + 7 = 0$ , is

1

\sqrt{2}

2 2

3 8

4 None of these

Explanation:

A Given,
$2 x^{2} + y^{2} - 8 x + 2 y + 7 = 0$
$2 x^{2} - 8 x + y^{2} + 2 y + 7 = 0$
$2 (x^{2} - 4 x + 4) - 8 + (y^{2} + 2 y + 1) - 1 + 7 = 0$
$2 (x - 2)^{2} + (y + 1)^{2} = 2$
$\frac{(x - 2)^{2}}{1} + \frac{(y + 1)^{2}}{2} = 1$ Hence, Length of latusrectum $= \frac{2 a^{2}}{b} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$

JCECE-2010

Ellipse

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
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$∴$ The general equation of the tangent of the ellipse is
$y = mx \pm \sqrt{a^{2} m^{2} + b^{2}}$
Now,
$3 x - 2 y - 20 = 0$
$∴ y = \frac{3}{2} x - 10$ is a tangent to the ellipse
Therefore, on comparing with equation (i), we get-
$m = \frac{3}{2} and a^{2} m^{2} + b^{2} = 100$
$a^{2} \cdot \frac{9}{4} + b^{2} = 100$
$9 a^{2} + 4 b^{2} = 400$
Similarly, line $x + 6 y - 20 = 0$
$y = - \frac{1}{6} x + \frac{10}{3}$
is a tangent to the ellipse is therefore on comparing with equation(i), we get :-
$m = - \frac{1}{6} and a^{2} m^{2} + b^{2} = \frac{100}{9}$
$\frac{a^{2}}{36} + b^{2} = \frac{100}{9}$
$a^{2} + 36 b^{2} = 400$
From equation (ii) and (iii), we get-
$a^{2} = 40 and b^{2} = 10$ Hence, $\frac{x^{2}}{40} + \frac{y^{2}}{10} = 1$ is required equations of ellipse.

JCECE-2010

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
$∴ 2 ae = 2$
$2 a = 10$
$a = 5$
$a^{2} = 25$
$∴ 2 \times 5 \times e = 2 e = \frac{1}{5}$
$∵ b^{2} = a^{2} (1 - e^{2}) b^{2} = 25 (1 - \frac{1}{25})$
$b^{2} = 24$
we know that, centre of ellipse $=$ mid point of foci
$= (\frac{2 + 4}{2}, \frac{2 + 2}{2}) = (3, 2)$
Hence, equation of ellipse is
$\frac{(x - 3)^{2}}{25} + \frac{(y - 2)^{2}}{24} = 1$

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 $(m_{1} \times m_{2}) = \frac{- b^{2}}{a^{2}}$
Where $m_{1}$ and $m_{2}$ are slopes of the respective diameters
$1 \times (\frac{- 2}{3}) = \frac{- b^{2}}{a^{2}}$
$b^{2} = \frac{2}{3} a^{2}$
How, $b^{2} = a^{2} (1 - e^{2}) = \frac{2}{3} a^{2} = a^{2} (1 - e^{2})$
$\frac{2}{3} = 1 - e^{2}$
$e^{2} = \frac{1}{3} = \frac{1}{\sqrt{3}}$

JCECE-2012

Ellipse

120519 The length of latusrectum of the ellipse $2 x^{2} + y^{2}$ $- 8 x + 2 y + 7 = 0$ , is

1

\sqrt{2}

2 2

3 8

4 None of these

Explanation:

A Given,
$2 x^{2} + y^{2} - 8 x + 2 y + 7 = 0$
$2 x^{2} - 8 x + y^{2} + 2 y + 7 = 0$
$2 (x^{2} - 4 x + 4) - 8 + (y^{2} + 2 y + 1) - 1 + 7 = 0$
$2 (x - 2)^{2} + (y + 1)^{2} = 2$
$\frac{(x - 2)^{2}}{1} + \frac{(y + 1)^{2}}{2} = 1$ Hence, Length of latusrectum $= \frac{2 a^{2}}{b} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$

JCECE-2010

Ellipse

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
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$∴$ The general equation of the tangent of the ellipse is
$y = mx \pm \sqrt{a^{2} m^{2} + b^{2}}$
Now,
$3 x - 2 y - 20 = 0$
$∴ y = \frac{3}{2} x - 10$ is a tangent to the ellipse
Therefore, on comparing with equation (i), we get-
$m = \frac{3}{2} and a^{2} m^{2} + b^{2} = 100$
$a^{2} \cdot \frac{9}{4} + b^{2} = 100$
$9 a^{2} + 4 b^{2} = 400$
Similarly, line $x + 6 y - 20 = 0$
$y = - \frac{1}{6} x + \frac{10}{3}$
is a tangent to the ellipse is therefore on comparing with equation(i), we get :-
$m = - \frac{1}{6} and a^{2} m^{2} + b^{2} = \frac{100}{9}$
$\frac{a^{2}}{36} + b^{2} = \frac{100}{9}$
$a^{2} + 36 b^{2} = 400$
From equation (ii) and (iii), we get-
$a^{2} = 40 and b^{2} = 10$ Hence, $\frac{x^{2}}{40} + \frac{y^{2}}{10} = 1$ is required equations of ellipse.

JCECE-2010

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
$∴ 2 ae = 2$
$2 a = 10$
$a = 5$
$a^{2} = 25$
$∴ 2 \times 5 \times e = 2 e = \frac{1}{5}$
$∵ b^{2} = a^{2} (1 - e^{2}) b^{2} = 25 (1 - \frac{1}{25})$
$b^{2} = 24$
we know that, centre of ellipse $=$ mid point of foci
$= (\frac{2 + 4}{2}, \frac{2 + 2}{2}) = (3, 2)$
Hence, equation of ellipse is
$\frac{(x - 3)^{2}}{25} + \frac{(y - 2)^{2}}{24} = 1$

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 $(m_{1} \times m_{2}) = \frac{- b^{2}}{a^{2}}$
Where $m_{1}$ and $m_{2}$ are slopes of the respective diameters
$1 \times (\frac{- 2}{3}) = \frac{- b^{2}}{a^{2}}$
$b^{2} = \frac{2}{3} a^{2}$
How, $b^{2} = a^{2} (1 - e^{2}) = \frac{2}{3} a^{2} = a^{2} (1 - e^{2})$
$\frac{2}{3} = 1 - e^{2}$
$e^{2} = \frac{1}{3} = \frac{1}{\sqrt{3}}$

JCECE-2012

Ellipse

120519 The length of latusrectum of the ellipse $2 x^{2} + y^{2}$ $- 8 x + 2 y + 7 = 0$ , is

1

\sqrt{2}

2 2

3 8

4 None of these

Explanation:

A Given,
$2 x^{2} + y^{2} - 8 x + 2 y + 7 = 0$
$2 x^{2} - 8 x + y^{2} + 2 y + 7 = 0$
$2 (x^{2} - 4 x + 4) - 8 + (y^{2} + 2 y + 1) - 1 + 7 = 0$
$2 (x - 2)^{2} + (y + 1)^{2} = 2$
$\frac{(x - 2)^{2}}{1} + \frac{(y + 1)^{2}}{2} = 1$ Hence, Length of latusrectum $= \frac{2 a^{2}}{b} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$

JCECE-2010

Ellipse

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
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$∴$ The general equation of the tangent of the ellipse is
$y = mx \pm \sqrt{a^{2} m^{2} + b^{2}}$
Now,
$3 x - 2 y - 20 = 0$
$∴ y = \frac{3}{2} x - 10$ is a tangent to the ellipse
Therefore, on comparing with equation (i), we get-
$m = \frac{3}{2} and a^{2} m^{2} + b^{2} = 100$
$a^{2} \cdot \frac{9}{4} + b^{2} = 100$
$9 a^{2} + 4 b^{2} = 400$
Similarly, line $x + 6 y - 20 = 0$
$y = - \frac{1}{6} x + \frac{10}{3}$
is a tangent to the ellipse is therefore on comparing with equation(i), we get :-
$m = - \frac{1}{6} and a^{2} m^{2} + b^{2} = \frac{100}{9}$
$\frac{a^{2}}{36} + b^{2} = \frac{100}{9}$
$a^{2} + 36 b^{2} = 400$
From equation (ii) and (iii), we get-
$a^{2} = 40 and b^{2} = 10$ Hence, $\frac{x^{2}}{40} + \frac{y^{2}}{10} = 1$ is required equations of ellipse.

JCECE-2010