Ellipse — Page 7 | JEE Mathematics

Q61

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus an its nearest vertex on the major axis is

{3 \over 2}

units, then its eccentricity is :

A

{1 \over 2}

B

{1 \over 3}

C

{2 \over 3}

D

{1 \over 9}

Correct Answer

Option B

Solution

If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively, then distance between focus and vertex, a $-$ ae =

{3 \over 2}

(given) $\Rightarrow$

\,\,\,

a (1 $-$ e) =

{3 \over 2}

Length of latus rectum,

{{2{b^2}} \over a} = 4

$\Rightarrow$

\,\,\,

b2 = 2a $\Rightarrow$

\,\,\,

a2(1 $-$ e2) = 2a [As b2 = a2 (1 $-$ e2)] $\Rightarrow$

\,\,\,

a (1 $-$ e) ( 1 + e) = 2 Putting a (1 $-$ e) =

{3 \over 2}

$\Rightarrow$

\,\,\,

{3 \over 2}

(1 + e) = 2 $\Rightarrow$

\,\,\,

3 + 3e = 4 $\Rightarrow$

\,\,\,

e =

{1 \over 3}

Q62

Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is

{3 \over 5}

and the distance between its foci is 6, then the area (in sq. units) of the quadrilatateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is :

A 8

B 32

C 80

D 40

Correct Answer

Option D

Solution

e = 3/5 & 2ae = 6 $\Rightarrow$ a = 5 $\because$ b2 = a2 (1 $-$ e2) $\Rightarrow$ b2 = 25(1 $-$ 9/25) $\Rightarrow$ b = 4 $\therefore$ Area of required quadrilateral = 4(1/2 ab) = 2ab = 40

Q63

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points (4, −1) and (−2, 2) is :

A

{1 \over 2}

B

{2 \over {\sqrt 5 }}

C

{{\sqrt 3 } \over 2}

D

{{\sqrt 3 } \over 4}

Correct Answer

Option C

Solution

Centre at (0, 0)

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}

= 1 at point (4, $-$ 1)

{{16} \over {{a^2}}} + {1 \over {{b^2}}}

= 1 $\Rightarrow$ 16b2 + a2 = a2b2 . . . .(i) at point ( $-$ 2, 2)

{4 \over {{a^2}}} + {4 \over {{b^2}}} = 1

$\Rightarrow$ 4b2 + 4a2 = a2b2 . . . .(ii) $\Rightarrow$ 16b2 + a2 = 4a2 + 4b2 From equations (i) and (ii) $\Rightarrow$ 3a2 = 12b2 $\Rightarrow$ a2 = 4b2 b2 = a2(1 $-$ e2) $\Rightarrow$ e2 =

{3 \over 4}

$\Rightarrow$ e =

{{\sqrt 3 } \over 2}

Q64

Let S and S' be the foci of an ellipse and B be any one of the extremities of its minor axis. If

\Delta

S'BS is a right angled triangle with right angle at B and area (

\Delta

S'BS) = 8 sq. units, then the length of a latus rectum of the ellipse is :

A 2

B 4

\sqrt 2

C 4

D 2

\sqrt 2

Correct Answer

Option C

Solution

b2 = a2e2 . . . . . . (i)

{1 \over 2}

S'B.SB = 8 S'B.SB = 16 a2e2 + b2 = 16 . . . . .(ii) b2 = a2 (1 $-$ e2) . . . . .(iii) using (i), (ii), (iii) a = 4 b =

2\sqrt 2

e =

{1 \over {\sqrt 2 }}

$\therefore$

\ell

(L.R) $=$

{{2{b^2}} \over a} = 4

Q65

Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

A

\left( {4\sqrt 2 ,2\sqrt 3 } \right)

B

\left( {4\sqrt 3 ,2\sqrt 3 } \right)

C

\left( {4\sqrt 3 ,2\sqrt 2 } \right)

D

\left( {4\sqrt 2 ,2\sqrt 2 } \right)

Correct Answer

Option C

Solution

{{2{b^2}} \over a} = 8

and 2ae $=$ 2b $\Rightarrow$

{b \over a}

= e and 1 $-$ e2 = e2 $\Rightarrow$ e $=$

{1 \over {\sqrt 2 }}

$\Rightarrow$ b = 4

\sqrt 2

and a $=$ 8 So equation of ellipse is

{{{x^2}} \over {64}} + {{{y^2}} \over {32}} = 1

Q66

Let S =

\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.

Then S represents :

A an ellipse whose eccentricity is

{1 \over {\sqrt {r + 1} }},

where r > 1

B an ellipse whose eccentricity is

{2 \over {\sqrt {r + 1} }},

where 0 < r < 1

C an ellipse whose eccentricity is

{2 \over {\sqrt {r - 1} }},

where 0 < r < 1

D an ellipse whose eccentricity is

\sqrt {{2 \over {r + 1}}}

, where r > 1

Correct Answer

Option D

Solution

{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}} = 1

for r > 1,

{{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1

e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)}

= \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}}

= \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}}

Q67

The eccentricity of an ellipse, with its centre at the origin, is

{1 \over 2}

. If one of the directrices is

x=4

, then the equation of the ellipse is :

A

4{x^2} + 3{y^2} = 1

B

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

C

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

D

3{x^2} + 4{y^2} = 1

Correct Answer

Option B

Solution

e = {1 \over 2}.\,\,

Directrix,

x = {a \over e} = 4

$\therefore$

a = 4 \times {1 \over 2} = 2

$\therefore$

b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3

Equation of elhipe is

{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12

Q68

An ellipse has

OB

as semi minor axis,

F

and

F

' its focii and theangle

FBF

' is a right angle. Then the eccentricity of the ellipse is :

A

{1 \over {\sqrt 2 }}

B

{1 \over 2}

C

{1 \over 4}

D

{1 \over {\sqrt 3 }}

Correct Answer

Option A

Solution

as

\angle FBF' = {90^ \circ }

\Rightarrow F{B^2} + F'{B^2} = FF{'^2}

$\therefore$

{\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + \left( {\sqrt {{a^2}{e^2} + {b^2}} } \right) = {\left( {2ae} \right)^2}

\Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2}

\Rightarrow {e^2} = {{{b^2}} \over {{a^2}}}

Also

{e^2} = 1 - {b^2}/{a^2} = 1 - {e^2}

\Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}

Q69

In the ellipse, the distance between its foci is

6

and minor axis is

8

. Then its eccentricity is :

A

{3 \over 5}

B

{1 \over 2}

C

{4 \over 5}

D

{1 \over {\sqrt 5 }}

Correct Answer

Option A

Solution

2ae = 6 \Rightarrow ae = 3;\,\,2b = 8 \Rightarrow b = 4

{b^2} = {a^2}\left( {1 - {e^2}} \right);16 = {a^2} - {a^2}{e^2}

\Rightarrow a{}^2 = 16 + 9 = 25

\Rightarrow a = 5

$\therefore$

e = {3 \over a} = {3 \over 5}

Q70

A focus of an ellipse is at the origin. The directrix is the line

x=4

and the eccentricity is

{{1 \over 2}}

. Then the length of the semi-major axis is :

A

{{8 \over 3}}

B

{{2 \over 3}}

C

{{4 \over 3}}

D

{{5 \over 3}}

Correct Answer

Option A

Solution

Perpendicular distance of directrix from focus

= {a \over e} - ae = 4

\Rightarrow a\left( {2 - {1 \over 2}} \right) = 4

\Rightarrow a = {8 \over 3}

$\therefore$ Semi major axis

=8/3