QBManager

Ellipse

120508 The eccentricity of the ellipse whose major axis is three times the minor axis is:

1

\frac{\sqrt{2}}{3}

2

\frac{\sqrt{3}}{2}

3

\frac{2 \sqrt{2}}{3}

4

\frac{2}{\sqrt{3}}

Explanation:

C Let,
$a =$ the major axis
$b =$ the minor axis of the ellipse,
3 minor axis $=$ major axis
$∴ 3 b = a$
Eccentricity is given by:
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = (3 b)^{2} (1 - e^{2})$
$b^{2} = 9 b^{2} (1 - e^{2})$
$\frac{1}{9} = 1 - e^{2}$
$\frac{1}{9} - 1 = - e^{2} = \frac{1 - 9}{9} \Rightarrow - e^{2}$
$- \frac{8}{9} = - e^{2}$
$e^{2} = \frac{8}{9}$
$e^{2} = \frac{\sqrt{8}}{3} or \frac{2 \sqrt{2}}{3}$

VITEEE-2018

Ellipse

120509 The foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$ coincide then value of $b^{2}$ is

1 1

2 5

3 7

4 9

Explanation:

C We have,
$\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$
Now,
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = 16 (1 - e^{2})$
$\frac{b^{2}}{16} = 1 - e^{2}$
$e^{2} = 1 - \frac{b^{2}}{16}$
$e^{2} = \frac{16 - b^{2}}{16}$
$e = \sqrt{\frac{16 - b^{2}}{16}}$
Foci $= (\pm ae, 0) \equiv \cdot (\pm \sqrt{16 - b^{2}}, 0)$
Given, hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$
On dividing 25 both sides,
$\frac{x^{2}}{{(\frac{12}{5})}^{2}} - \frac{y^{2}}{{(\frac{9}{5})}^{2}} = 1$
Now,
${(\frac{9}{5})}^{2} = {(\frac{12}{5})}^{2} (e^{2} - 1)$
${(\frac{9}{12})}^{2} = (e^{2} - 1)$
$e^{2} - 1 = \frac{81}{144}$
$e^{2} = \frac{81}{144} + 1$
$e^{2} = \frac{81 + 144}{144} = \frac{225}{144}$
$e^{2} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$
So, Foci $= (\pm$ ae, 0 $) = (\pm 3, 0)$
Since foci of the given ellipse and hyperbola coincide, therefore
$\sqrt{16 - b^{2}} = 3$
$16 - b^{2} = 9$
$b^{2} = 16 - 9$
$∴ b^{2} = 7$

VITEEE-2016

Ellipse

120510 The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with the coordinate axes is

1

a^{2} + b^{2}

2

\frac{(a + b)^{2}}{2}

3 ab

4

\frac{(a - b)^{2}}{2}

Explanation:

C We know that, equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Equation of tangent at $(a \cos θ, b \sin θ)$ to the ellipse is
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
original image
Coordinates of $P$ and $Q$ are $(\frac{a}{\cos θ}, 0)$ and $(0, \frac{b}{\sin θ})$ , respectively.
Area of $Δ OPQ = \frac{1}{2} | \frac{a}{\cos θ} \times \frac{b}{\sin θ} |$
$= \frac{ab}{| \sin 2 θ |}$ $∴$ Minimum area $ab$

VITEEE-2014

Ellipse

120511 If the equation of an ellipse is $3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$ , then which of the following are true?

1

e = \frac{1}{\sqrt{3}}

2 center is

(- 1, 2)

3 foci are

(- 1, 1)

and

(- 1, 3)

4 All of the above

Explanation:

D
The equation of ellipse is
$3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$
$3 (x^{2} + 2 x) + 2 (y^{2} - 4 y) + 5 = 0$
$3 (x^{2} + 2 x + 1) + 2 (y^{2} - 4 y + 4) + 5 - 3 - 8 = 0$
$3 (x + 1)^{2} + 2 (y - 2)^{2} = 6$
$\frac{(x + 1)^{2}}{2} + \frac{(y - 2)^{2}}{3} = 1$
Comparing with
$\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1, we get -$
$h = - 1, k = 2, a^{2} = 2, b^{2} = 3$
Here, centre $(h, k) = (- 1, 2)$
And using $a^{2} = b^{2} (1 - e^{2})$
$2 = 3 (1 - e^{2}) \Rightarrow e = \frac{1}{\sqrt{3}}$
And foci are ( $h, k + be)$ and ( $h, k - be)$
$= (- 1, 2 + 1)$ and $(- 1, 2 - 1) (\binom{∵ b e = \sqrt{3} \times \frac{1}{\sqrt{3}}}{b e = 1})$
$= (- 1, 3)$ and $(- 1, 1)$

VITEEE-2010

Ellipse

120512 The equation of an ellipse with focus at $(1, - 1)$ , directrix $x - y - 3 = 0$ and eccentricity $1 / 2$ is

1

7 x^{2} + 2 x y + 7 y^{2} + 7 = 0

2

7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0

3

7 (x^{2} + y^{2}) + 2 x y + 10 x - 10 y - 7 = 0

4

7 (x^{2} + y^{2}) + 2 x y - 10 x - 10 y + 7 = 0

Explanation:

B Given that,
Equation of directrix is
$x - y = 3$
Focus $= (1, - 1)$ .
Eccentricity (e) $= \frac{1}{2}$
We know that, perpendicular distance from point ( $x_{1}$ , $y_{1}$ ) to the line $ax + by + c = 0$ is
$\frac{| {ax}_{1} + {by}_{1} + c |}{\sqrt{a^{2} + b^{2}}}$
$∴ (SP)^{2} = (ePM)^{2}$
${SP}^{2} = {(\frac{1}{2})}^{2} {PM}^{2}$
$4 {SP}^{2} = {PM}^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = (x - y - 3)^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x$
i.e. $4 [(x - 1)^{2} + (y + 1)^{2}] = {(\frac{x - y - 3}{\sqrt{1^{2} + (- 1)^{2}}})}^{2}$
On simplifying, we get-
$7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0$

UPSEE-2018

Ellipse

120508 The eccentricity of the ellipse whose major axis is three times the minor axis is:

1

\frac{\sqrt{2}}{3}

2

\frac{\sqrt{3}}{2}

3

\frac{2 \sqrt{2}}{3}

4

\frac{2}{\sqrt{3}}

Explanation:

C Let,
$a =$ the major axis
$b =$ the minor axis of the ellipse,
3 minor axis $=$ major axis
$∴ 3 b = a$
Eccentricity is given by:
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = (3 b)^{2} (1 - e^{2})$
$b^{2} = 9 b^{2} (1 - e^{2})$
$\frac{1}{9} = 1 - e^{2}$
$\frac{1}{9} - 1 = - e^{2} = \frac{1 - 9}{9} \Rightarrow - e^{2}$
$- \frac{8}{9} = - e^{2}$
$e^{2} = \frac{8}{9}$
$e^{2} = \frac{\sqrt{8}}{3} or \frac{2 \sqrt{2}}{3}$

VITEEE-2018

Ellipse

120509 The foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$ coincide then value of $b^{2}$ is

1 1

2 5

3 7

4 9

Explanation:

C We have,
$\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$
Now,
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = 16 (1 - e^{2})$
$\frac{b^{2}}{16} = 1 - e^{2}$
$e^{2} = 1 - \frac{b^{2}}{16}$
$e^{2} = \frac{16 - b^{2}}{16}$
$e = \sqrt{\frac{16 - b^{2}}{16}}$
Foci $= (\pm ae, 0) \equiv \cdot (\pm \sqrt{16 - b^{2}}, 0)$
Given, hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$
On dividing 25 both sides,
$\frac{x^{2}}{{(\frac{12}{5})}^{2}} - \frac{y^{2}}{{(\frac{9}{5})}^{2}} = 1$
Now,
${(\frac{9}{5})}^{2} = {(\frac{12}{5})}^{2} (e^{2} - 1)$
${(\frac{9}{12})}^{2} = (e^{2} - 1)$
$e^{2} - 1 = \frac{81}{144}$
$e^{2} = \frac{81}{144} + 1$
$e^{2} = \frac{81 + 144}{144} = \frac{225}{144}$
$e^{2} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$
So, Foci $= (\pm$ ae, 0 $) = (\pm 3, 0)$
Since foci of the given ellipse and hyperbola coincide, therefore
$\sqrt{16 - b^{2}} = 3$
$16 - b^{2} = 9$
$b^{2} = 16 - 9$
$∴ b^{2} = 7$

VITEEE-2016

Ellipse

120510 The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with the coordinate axes is

1

a^{2} + b^{2}

2

\frac{(a + b)^{2}}{2}

3 ab

4

\frac{(a - b)^{2}}{2}

Explanation:

C We know that, equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Equation of tangent at $(a \cos θ, b \sin θ)$ to the ellipse is
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
original image
Coordinates of $P$ and $Q$ are $(\frac{a}{\cos θ}, 0)$ and $(0, \frac{b}{\sin θ})$ , respectively.
Area of $Δ OPQ = \frac{1}{2} | \frac{a}{\cos θ} \times \frac{b}{\sin θ} |$
$= \frac{ab}{| \sin 2 θ |}$ $∴$ Minimum area $ab$

VITEEE-2014

Ellipse

120511 If the equation of an ellipse is $3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$ , then which of the following are true?

1

e = \frac{1}{\sqrt{3}}

2 center is

(- 1, 2)

3 foci are

(- 1, 1)

and

(- 1, 3)

4 All of the above

Explanation:

D
The equation of ellipse is
$3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$
$3 (x^{2} + 2 x) + 2 (y^{2} - 4 y) + 5 = 0$
$3 (x^{2} + 2 x + 1) + 2 (y^{2} - 4 y + 4) + 5 - 3 - 8 = 0$
$3 (x + 1)^{2} + 2 (y - 2)^{2} = 6$
$\frac{(x + 1)^{2}}{2} + \frac{(y - 2)^{2}}{3} = 1$
Comparing with
$\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1, we get -$
$h = - 1, k = 2, a^{2} = 2, b^{2} = 3$
Here, centre $(h, k) = (- 1, 2)$
And using $a^{2} = b^{2} (1 - e^{2})$
$2 = 3 (1 - e^{2}) \Rightarrow e = \frac{1}{\sqrt{3}}$
And foci are ( $h, k + be)$ and ( $h, k - be)$
$= (- 1, 2 + 1)$ and $(- 1, 2 - 1) (\binom{∵ b e = \sqrt{3} \times \frac{1}{\sqrt{3}}}{b e = 1})$
$= (- 1, 3)$ and $(- 1, 1)$

VITEEE-2010

Ellipse

120512 The equation of an ellipse with focus at $(1, - 1)$ , directrix $x - y - 3 = 0$ and eccentricity $1 / 2$ is

1

7 x^{2} + 2 x y + 7 y^{2} + 7 = 0

2

7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0

3

7 (x^{2} + y^{2}) + 2 x y + 10 x - 10 y - 7 = 0

4

7 (x^{2} + y^{2}) + 2 x y - 10 x - 10 y + 7 = 0

Explanation:

B Given that,
Equation of directrix is
$x - y = 3$
Focus $= (1, - 1)$ .
Eccentricity (e) $= \frac{1}{2}$
We know that, perpendicular distance from point ( $x_{1}$ , $y_{1}$ ) to the line $ax + by + c = 0$ is
$\frac{| {ax}_{1} + {by}_{1} + c |}{\sqrt{a^{2} + b^{2}}}$
$∴ (SP)^{2} = (ePM)^{2}$
${SP}^{2} = {(\frac{1}{2})}^{2} {PM}^{2}$
$4 {SP}^{2} = {PM}^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = (x - y - 3)^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x$
i.e. $4 [(x - 1)^{2} + (y + 1)^{2}] = {(\frac{x - y - 3}{\sqrt{1^{2} + (- 1)^{2}}})}^{2}$
On simplifying, we get-
$7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0$

UPSEE-2018

Ellipse

120508 The eccentricity of the ellipse whose major axis is three times the minor axis is:

1

\frac{\sqrt{2}}{3}

2

\frac{\sqrt{3}}{2}

3

\frac{2 \sqrt{2}}{3}

4

\frac{2}{\sqrt{3}}

Explanation:

C Let,
$a =$ the major axis
$b =$ the minor axis of the ellipse,
3 minor axis $=$ major axis
$∴ 3 b = a$
Eccentricity is given by:
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = (3 b)^{2} (1 - e^{2})$
$b^{2} = 9 b^{2} (1 - e^{2})$
$\frac{1}{9} = 1 - e^{2}$
$\frac{1}{9} - 1 = - e^{2} = \frac{1 - 9}{9} \Rightarrow - e^{2}$
$- \frac{8}{9} = - e^{2}$
$e^{2} = \frac{8}{9}$
$e^{2} = \frac{\sqrt{8}}{3} or \frac{2 \sqrt{2}}{3}$

VITEEE-2018

Ellipse

120509 The foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$ coincide then value of $b^{2}$ is

1 1

2 5

3 7

4 9

Explanation:

C We have,
$\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$
Now,
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = 16 (1 - e^{2})$
$\frac{b^{2}}{16} = 1 - e^{2}$
$e^{2} = 1 - \frac{b^{2}}{16}$
$e^{2} = \frac{16 - b^{2}}{16}$
$e = \sqrt{\frac{16 - b^{2}}{16}}$
Foci $= (\pm ae, 0) \equiv \cdot (\pm \sqrt{16 - b^{2}}, 0)$
Given, hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$
On dividing 25 both sides,
$\frac{x^{2}}{{(\frac{12}{5})}^{2}} - \frac{y^{2}}{{(\frac{9}{5})}^{2}} = 1$
Now,
${(\frac{9}{5})}^{2} = {(\frac{12}{5})}^{2} (e^{2} - 1)$
${(\frac{9}{12})}^{2} = (e^{2} - 1)$
$e^{2} - 1 = \frac{81}{144}$
$e^{2} = \frac{81}{144} + 1$
$e^{2} = \frac{81 + 144}{144} = \frac{225}{144}$
$e^{2} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$
So, Foci $= (\pm$ ae, 0 $) = (\pm 3, 0)$
Since foci of the given ellipse and hyperbola coincide, therefore
$\sqrt{16 - b^{2}} = 3$
$16 - b^{2} = 9$
$b^{2} = 16 - 9$
$∴ b^{2} = 7$

VITEEE-2016

Ellipse

120510 The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with the coordinate axes is

1

a^{2} + b^{2}

2

\frac{(a + b)^{2}}{2}

3 ab

4

\frac{(a - b)^{2}}{2}

Explanation:

C We know that, equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Equation of tangent at $(a \cos θ, b \sin θ)$ to the ellipse is
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
original image
Coordinates of $P$ and $Q$ are $(\frac{a}{\cos θ}, 0)$ and $(0, \frac{b}{\sin θ})$ , respectively.
Area of $Δ OPQ = \frac{1}{2} | \frac{a}{\cos θ} \times \frac{b}{\sin θ} |$
$= \frac{ab}{| \sin 2 θ |}$ $∴$ Minimum area $ab$

VITEEE-2014

Ellipse

120511 If the equation of an ellipse is $3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$ , then which of the following are true?

1

e = \frac{1}{\sqrt{3}}

2 center is

(- 1, 2)

3 foci are

(- 1, 1)

and

(- 1, 3)

4 All of the above

Explanation:

D
The equation of ellipse is
$3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$
$3 (x^{2} + 2 x) + 2 (y^{2} - 4 y) + 5 = 0$
$3 (x^{2} + 2 x + 1) + 2 (y^{2} - 4 y + 4) + 5 - 3 - 8 = 0$
$3 (x + 1)^{2} + 2 (y - 2)^{2} = 6$
$\frac{(x + 1)^{2}}{2} + \frac{(y - 2)^{2}}{3} = 1$
Comparing with
$\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1, we get -$
$h = - 1, k = 2, a^{2} = 2, b^{2} = 3$
Here, centre $(h, k) = (- 1, 2)$
And using $a^{2} = b^{2} (1 - e^{2})$
$2 = 3 (1 - e^{2}) \Rightarrow e = \frac{1}{\sqrt{3}}$
And foci are ( $h, k + be)$ and ( $h, k - be)$
$= (- 1, 2 + 1)$ and $(- 1, 2 - 1) (\binom{∵ b e = \sqrt{3} \times \frac{1}{\sqrt{3}}}{b e = 1})$
$= (- 1, 3)$ and $(- 1, 1)$

VITEEE-2010

Ellipse

120512 The equation of an ellipse with focus at $(1, - 1)$ , directrix $x - y - 3 = 0$ and eccentricity $1 / 2$ is

1

7 x^{2} + 2 x y + 7 y^{2} + 7 = 0

2

7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0

3

7 (x^{2} + y^{2}) + 2 x y + 10 x - 10 y - 7 = 0

4

7 (x^{2} + y^{2}) + 2 x y - 10 x - 10 y + 7 = 0

Explanation:

B Given that,
Equation of directrix is
$x - y = 3$
Focus $= (1, - 1)$ .
Eccentricity (e) $= \frac{1}{2}$
We know that, perpendicular distance from point ( $x_{1}$ , $y_{1}$ ) to the line $ax + by + c = 0$ is
$\frac{| {ax}_{1} + {by}_{1} + c |}{\sqrt{a^{2} + b^{2}}}$
$∴ (SP)^{2} = (ePM)^{2}$
${SP}^{2} = {(\frac{1}{2})}^{2} {PM}^{2}$
$4 {SP}^{2} = {PM}^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = (x - y - 3)^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x$
i.e. $4 [(x - 1)^{2} + (y + 1)^{2}] = {(\frac{x - y - 3}{\sqrt{1^{2} + (- 1)^{2}}})}^{2}$
On simplifying, we get-
$7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0$

UPSEE-2018

Ellipse

120508 The eccentricity of the ellipse whose major axis is three times the minor axis is:

1

\frac{\sqrt{2}}{3}

2

\frac{\sqrt{3}}{2}

3

\frac{2 \sqrt{2}}{3}

4

\frac{2}{\sqrt{3}}

Explanation:

C Let,
$a =$ the major axis
$b =$ the minor axis of the ellipse,
3 minor axis $=$ major axis
$∴ 3 b = a$
Eccentricity is given by:
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = (3 b)^{2} (1 - e^{2})$
$b^{2} = 9 b^{2} (1 - e^{2})$
$\frac{1}{9} = 1 - e^{2}$
$\frac{1}{9} - 1 = - e^{2} = \frac{1 - 9}{9} \Rightarrow - e^{2}$
$- \frac{8}{9} = - e^{2}$
$e^{2} = \frac{8}{9}$
$e^{2} = \frac{\sqrt{8}}{3} or \frac{2 \sqrt{2}}{3}$

VITEEE-2018

Ellipse

120509 The foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$ coincide then value of $b^{2}$ is

1 1

2 5

3 7

4 9

Explanation:

C We have,
$\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$
Now,
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = 16 (1 - e^{2})$
$\frac{b^{2}}{16} = 1 - e^{2}$
$e^{2} = 1 - \frac{b^{2}}{16}$
$e^{2} = \frac{16 - b^{2}}{16}$
$e = \sqrt{\frac{16 - b^{2}}{16}}$
Foci $= (\pm ae, 0) \equiv \cdot (\pm \sqrt{16 - b^{2}}, 0)$
Given, hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$
On dividing 25 both sides,
$\frac{x^{2}}{{(\frac{12}{5})}^{2}} - \frac{y^{2}}{{(\frac{9}{5})}^{2}} = 1$
Now,
${(\frac{9}{5})}^{2} = {(\frac{12}{5})}^{2} (e^{2} - 1)$
${(\frac{9}{12})}^{2} = (e^{2} - 1)$
$e^{2} - 1 = \frac{81}{144}$
$e^{2} = \frac{81}{144} + 1$
$e^{2} = \frac{81 + 144}{144} = \frac{225}{144}$
$e^{2} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$
So, Foci $= (\pm$ ae, 0 $) = (\pm 3, 0)$
Since foci of the given ellipse and hyperbola coincide, therefore
$\sqrt{16 - b^{2}} = 3$
$16 - b^{2} = 9$
$b^{2} = 16 - 9$
$∴ b^{2} = 7$

VITEEE-2016

Ellipse

120510 The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with the coordinate axes is

1

a^{2} + b^{2}

2

\frac{(a + b)^{2}}{2}

3 ab

4

\frac{(a - b)^{2}}{2}

Explanation:

C We know that, equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Equation of tangent at $(a \cos θ, b \sin θ)$ to the ellipse is
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
original image
Coordinates of $P$ and $Q$ are $(\frac{a}{\cos θ}, 0)$ and $(0, \frac{b}{\sin θ})$ , respectively.
Area of $Δ OPQ = \frac{1}{2} | \frac{a}{\cos θ} \times \frac{b}{\sin θ} |$
$= \frac{ab}{| \sin 2 θ |}$ $∴$ Minimum area $ab$

VITEEE-2014

Ellipse

120511 If the equation of an ellipse is $3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$ , then which of the following are true?

1

e = \frac{1}{\sqrt{3}}

2 center is

(- 1, 2)

3 foci are

(- 1, 1)

and

(- 1, 3)

4 All of the above

Explanation:

D
The equation of ellipse is
$3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$
$3 (x^{2} + 2 x) + 2 (y^{2} - 4 y) + 5 = 0$
$3 (x^{2} + 2 x + 1) + 2 (y^{2} - 4 y + 4) + 5 - 3 - 8 = 0$
$3 (x + 1)^{2} + 2 (y - 2)^{2} = 6$
$\frac{(x + 1)^{2}}{2} + \frac{(y - 2)^{2}}{3} = 1$
Comparing with
$\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1, we get -$
$h = - 1, k = 2, a^{2} = 2, b^{2} = 3$
Here, centre $(h, k) = (- 1, 2)$
And using $a^{2} = b^{2} (1 - e^{2})$
$2 = 3 (1 - e^{2}) \Rightarrow e = \frac{1}{\sqrt{3}}$
And foci are ( $h, k + be)$ and ( $h, k - be)$
$= (- 1, 2 + 1)$ and $(- 1, 2 - 1) (\binom{∵ b e = \sqrt{3} \times \frac{1}{\sqrt{3}}}{b e = 1})$
$= (- 1, 3)$ and $(- 1, 1)$

VITEEE-2010

Ellipse

120512 The equation of an ellipse with focus at $(1, - 1)$ , directrix $x - y - 3 = 0$ and eccentricity $1 / 2$ is

1

7 x^{2} + 2 x y + 7 y^{2} + 7 = 0

2

7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0

3

7 (x^{2} + y^{2}) + 2 x y + 10 x - 10 y - 7 = 0

4

7 (x^{2} + y^{2}) + 2 x y - 10 x - 10 y + 7 = 0

Explanation:

B Given that,
Equation of directrix is
$x - y = 3$
Focus $= (1, - 1)$ .
Eccentricity (e) $= \frac{1}{2}$
We know that, perpendicular distance from point ( $x_{1}$ , $y_{1}$ ) to the line $ax + by + c = 0$ is
$\frac{| {ax}_{1} + {by}_{1} + c |}{\sqrt{a^{2} + b^{2}}}$
$∴ (SP)^{2} = (ePM)^{2}$
${SP}^{2} = {(\frac{1}{2})}^{2} {PM}^{2}$
$4 {SP}^{2} = {PM}^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = (x - y - 3)^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x$
i.e. $4 [(x - 1)^{2} + (y + 1)^{2}] = {(\frac{x - y - 3}{\sqrt{1^{2} + (- 1)^{2}}})}^{2}$
On simplifying, we get-
$7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0$

UPSEE-2018

Ellipse

120508 The eccentricity of the ellipse whose major axis is three times the minor axis is:

1

\frac{\sqrt{2}}{3}

2

\frac{\sqrt{3}}{2}

3

\frac{2 \sqrt{2}}{3}

4

\frac{2}{\sqrt{3}}

Explanation:

C Let,
$a =$ the major axis
$b =$ the minor axis of the ellipse,
3 minor axis $=$ major axis
$∴ 3 b = a$
Eccentricity is given by:
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = (3 b)^{2} (1 - e^{2})$
$b^{2} = 9 b^{2} (1 - e^{2})$
$\frac{1}{9} = 1 - e^{2}$
$\frac{1}{9} - 1 = - e^{2} = \frac{1 - 9}{9} \Rightarrow - e^{2}$
$- \frac{8}{9} = - e^{2}$
$e^{2} = \frac{8}{9}$
$e^{2} = \frac{\sqrt{8}}{3} or \frac{2 \sqrt{2}}{3}$

VITEEE-2018

Ellipse

120509 The foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$ coincide then value of $b^{2}$ is

1 1

2 5

3 7

4 9

Explanation:

C We have,
$\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$
Now,
$b^{2} = a^{2} (1 - e^{2})$
$b^{2} = 16 (1 - e^{2})$
$\frac{b^{2}}{16} = 1 - e^{2}$
$e^{2} = 1 - \frac{b^{2}}{16}$
$e^{2} = \frac{16 - b^{2}}{16}$
$e = \sqrt{\frac{16 - b^{2}}{16}}$
Foci $= (\pm ae, 0) \equiv \cdot (\pm \sqrt{16 - b^{2}}, 0)$
Given, hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$
On dividing 25 both sides,
$\frac{x^{2}}{{(\frac{12}{5})}^{2}} - \frac{y^{2}}{{(\frac{9}{5})}^{2}} = 1$
Now,
${(\frac{9}{5})}^{2} = {(\frac{12}{5})}^{2} (e^{2} - 1)$
${(\frac{9}{12})}^{2} = (e^{2} - 1)$
$e^{2} - 1 = \frac{81}{144}$
$e^{2} = \frac{81}{144} + 1$
$e^{2} = \frac{81 + 144}{144} = \frac{225}{144}$
$e^{2} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$
So, Foci $= (\pm$ ae, 0 $) = (\pm 3, 0)$
Since foci of the given ellipse and hyperbola coincide, therefore
$\sqrt{16 - b^{2}} = 3$
$16 - b^{2} = 9$
$b^{2} = 16 - 9$
$∴ b^{2} = 7$

VITEEE-2016

Ellipse

120510 The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with the coordinate axes is

1

a^{2} + b^{2}

2

\frac{(a + b)^{2}}{2}

3 ab

4

\frac{(a - b)^{2}}{2}

Explanation:

C We know that, equation of ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Equation of tangent at $(a \cos θ, b \sin θ)$ to the ellipse is
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
original image
Coordinates of $P$ and $Q$ are $(\frac{a}{\cos θ}, 0)$ and $(0, \frac{b}{\sin θ})$ , respectively.
Area of $Δ OPQ = \frac{1}{2} | \frac{a}{\cos θ} \times \frac{b}{\sin θ} |$
$= \frac{ab}{| \sin 2 θ |}$ $∴$ Minimum area $ab$

VITEEE-2014

Ellipse

120511 If the equation of an ellipse is $3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$ , then which of the following are true?

1

e = \frac{1}{\sqrt{3}}

2 center is

(- 1, 2)

3 foci are

(- 1, 1)

and

(- 1, 3)

4 All of the above

Explanation:

D
The equation of ellipse is
$3 x^{2} + 2 y^{2} + 6 x - 8 y + 5 = 0$
$3 (x^{2} + 2 x) + 2 (y^{2} - 4 y) + 5 = 0$
$3 (x^{2} + 2 x + 1) + 2 (y^{2} - 4 y + 4) + 5 - 3 - 8 = 0$
$3 (x + 1)^{2} + 2 (y - 2)^{2} = 6$
$\frac{(x + 1)^{2}}{2} + \frac{(y - 2)^{2}}{3} = 1$
Comparing with
$\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1, we get -$
$h = - 1, k = 2, a^{2} = 2, b^{2} = 3$
Here, centre $(h, k) = (- 1, 2)$
And using $a^{2} = b^{2} (1 - e^{2})$
$2 = 3 (1 - e^{2}) \Rightarrow e = \frac{1}{\sqrt{3}}$
And foci are ( $h, k + be)$ and ( $h, k - be)$
$= (- 1, 2 + 1)$ and $(- 1, 2 - 1) (\binom{∵ b e = \sqrt{3} \times \frac{1}{\sqrt{3}}}{b e = 1})$
$= (- 1, 3)$ and $(- 1, 1)$

VITEEE-2010

Ellipse

120512 The equation of an ellipse with focus at $(1, - 1)$ , directrix $x - y - 3 = 0$ and eccentricity $1 / 2$ is

1

7 x^{2} + 2 x y + 7 y^{2} + 7 = 0

2

7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0

3

7 (x^{2} + y^{2}) + 2 x y + 10 x - 10 y - 7 = 0

4

7 (x^{2} + y^{2}) + 2 x y - 10 x - 10 y + 7 = 0

Explanation:

B Given that,
Equation of directrix is
$x - y = 3$
Focus $= (1, - 1)$ .
Eccentricity (e) $= \frac{1}{2}$
We know that, perpendicular distance from point ( $x_{1}$ , $y_{1}$ ) to the line $ax + by + c = 0$ is
$\frac{| {ax}_{1} + {by}_{1} + c |}{\sqrt{a^{2} + b^{2}}}$
$∴ (SP)^{2} = (ePM)^{2}$
${SP}^{2} = {(\frac{1}{2})}^{2} {PM}^{2}$
$4 {SP}^{2} = {PM}^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = (x - y - 3)^{2}$
$8 (x^{2} + y^{2} - 2 x + 2 y + 2) = x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x$
i.e. $4 [(x - 1)^{2} + (y + 1)^{2}] = {(\frac{x - y - 3}{\sqrt{1^{2} + (- 1)^{2}}})}^{2}$
On simplifying, we get-
$7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0$

UPSEE-2018

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