Ellipse — Page 8 | JEE Mathematics

Q71

The ellipse

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

is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point

(4,0)

. Then the equation of the ellipse is :

A

{x^2} + 12{y^2} = 16

B

4{x^2} + 48{y^2} = 48

C

4{x^2} + 64{y^2} = 48

D

{x^2} + 16{y^2} = 16

Correct Answer

Option A

Solution

The given ellipse is

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

So

A=(2,0)

and

B = \left( {0,1} \right)

If

PQRS

is the rectangular in which it is inscribed, then

P = \left( {2,1} \right).

Let

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

be the ellipse circumscribing the rectangular

PQRS

. Then it passes through

P\,\,(2,1)

$\therefore$

{4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)

Also, given that, it passes through

(4,0)

$\therefore$

{{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16

\Rightarrow {b^2} = 4/3

\left[ {\,\,} \right.

substituting

{{a^2} = 16\,\,}

in

\left. {e{q^n}\left( a \right)\,\,} \right]

$\therefore$ The required ellipse is

{{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1

or

{x^2} + 12y{}^2 = 16

Q72

Equation of the ellipse whose axes of coordinates and which passes through the point

(-3,1)

and has eccentricity

\sqrt {{2 \over 5}}

is :

A

5{x^2} + 3{y^2} - 48 = 0

B

3{x^2} + 5{y^2} - 15 = 0

C

5{x^2} + 3{y^2} - 32 = 0

D

3{x^2} + 5{y^2} - 32 = 0

Correct Answer

Option D

Solution

Let the ellipse be

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

It press through

(-3, 1)

so

{9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)

Also,

{b^2} = {a^2}\left( {1 - 2/5} \right)

\Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

Solving

(i)

and

(ii)

we get

{a^2} = {{32} \over 3},{b^2} = {{32} \over 5}

So, the equation of the ellipse is

3{x^2} + 5{y^2} = 32

Q73

STATEMENT-1 : An equation of a common tangent to the parabola

{y^2} = 16\sqrt 3 x

and the ellipse

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

is

y = 2x + 2\sqrt 3

STATEMENT-2 :If line

y = mx + {{4\sqrt 3 } \over m},\left( {m \ne 0} \right)

is a common tangent to the parabola

{y^2} = 16\sqrt {3x}

and the ellipse

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

, then

m

satisfies

{m^4} + 2{m^2} = 24

A Statement-1 is false, Statement-2 is true.

B Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

C Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

D Statement-1 is true, Statement-2 is false.

Correct Answer

Option B

Solution

Given equation of ellipse is

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

\Rightarrow {{2{x^2}} \over 4} + {{{y^2}} \over 4} = 1 \Rightarrow {{{x_2}} \over 2} + {{{y^2}} \over 4} = 1

Equation of tangent to the ellipse

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

is

y = mx \pm \sqrt {2{m^2} + 4} \,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

( as equation of tangent to the ellipse

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

is

y=mx+c

where

c = \pm \sqrt {{a^2}{m^2} + {b^2}}

) Now, Equation of tangent to the parabola

{y^2} = 16\sqrt 3 x

is

y = mx + {{4\sqrt 3 } \over m}\,\,\,\,\,\,\,\,...\left( 2 \right)

( as equation of tangent to the parabola

{y^2} = 4ax

is

y = mx + {a \over m}

) On comparing

(1)

and

(2),

we get

{{4\sqrt 3 } \over m} = \pm \sqrt {2{m^2} + 4}

Squaring on both the sides, we get

16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2}

\Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \Rightarrow 2{m^4} + 4{m^2} - 48 = 0

\Rightarrow {m^4} + 2{m^2} - 24 = 0 \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0

\Rightarrow {m^2} = 4

( as

{m^2} \ne - 6

)

\Rightarrow m = \pm 2

$\Rightarrow$ Equation of common tangents are

y = \pm 2x \pm 2\sqrt 3

Thus, statement -

1

is true. Statement -

2

is obviously true.

Q74

An ellipse is drawn by taking a diameter of thec circle

{\left( {x - 1} \right)^2} + {y^2} = 1

as its semi-minor axis and a diameter of the circle

{x^2} + {\left( {y - 2} \right)^2} = 4

is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :

A

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

B

{x^2} + 4{y^2} = 8

C

4{x^2} + {y^2} = 8

D

{x^2} + 4{y^2} = 16

Correct Answer

Option D

Solution

Equation of circle is

{\left( {x - 1} \right)^2} + {y^2} = 1

$\Rightarrow$ radius

=1

and diameter

=2

$\therefore$ Length of semi-minor axis is

2.

Equation of circle is

{x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}

$\Rightarrow$ radius

=2

and diameter

=4

$\therefore$ Length of semi major axis is

4

We know, equation of ellipse is given by

{{{x^2}} \over {\left( {Major\,\,\,axi{s^{\,\,2}}} \right)}} + {{{y^2}} \over {\left( {Minor\,\,\,axi{s^{\,\,2}}} \right)}} = 1

\Rightarrow {{{x^2}} \over {{{\left( 4 \right)}^2}}} + {{{y^2}} \over {{{\left( 2 \right)}^2}}} = 1

\Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 4} = 1

\Rightarrow {x^2} + 4{y^2} = 16

Q75

The equation of the circle passing through the foci of the ellipse

{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1

, and having centre at

(0,3)

is :

A

{x^2} + {y^2} - 6y - 7 = 0

B

{x^2} + {y^2} - 6y + 7 = 0

C

{x^2} + {y^2} - 6y - 5 = 0

D

{x^2} + {y^2} - 6y + 5 = 0

Correct Answer

Option A

Solution

From the given equation of ellipse, we have

a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}}

\Rightarrow e = {{\sqrt 7 } \over 4}

Now, radius of this circle

= {a^2} = 16

\Rightarrow Focii = \left( { \pm \sqrt 7 ,0} \right)

Now equation of circle is

{\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16

{x^2} + y{}^2 - 6y - 7 = 0

Q76

The locus of the foot of perpendicular drawn from the centre of the ellipse

{x^2} + 3{y^2} = 6

on any tangent to it is :

A

\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} + 2{y^2}

B

\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} - 2{y^2}

C

\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} + 2{y^2}

D

\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} - 2{y^2}

Correct Answer

Option A

Solution

Given

e{q^n}

of ellipse can be written as

{{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2

Now, equation of any variable tangent is

y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)

where

m

is slope of the tangent So, equation of perpendicular line drawn- from center to tangent is

y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

Eliminating

m,

we get

\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}

\Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}

\Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}

Q77

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse

{{{x^2}} \over 9} + {{{y^2}} \over 5} = 1

, is :

A

{{27 \over 2}}

B

27

C

{{27 \over 4}}

D

18

Correct Answer

Option B

Solution

The end point of latus rectum of ellipse

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

in first quadrant is

\left( {ae,{{{b^2}} \over a}} \right)

and the tangent at this point intersects

x

-axis at

\left( {{a \over e},0} \right)

and

y

-axis at

(0,a).

The given ellipse is

{{x{}^2} \over 9} + {{{y^2}} \over 5} = 1

Then

{a^2} = 9,{b^2} = 5

\Rightarrow e = \sqrt {1 - {5 \over 9}} = {2 \over 3}

$\therefore$ end point of latus rectum in first quadrant is

L\left( {2,\,\,5/3} \right)

Equation of tangent at

L

is

{{2x} \over 9} + {y \over 3} - 1

It meets

x

-axis at

A(9/2, 0)

and

y

-axis at

B(0,3)

$\therefore$ Area of

\Delta OAB = {1 \over 2} \times {9 \over 2} \times 3 = {{27} \over 4}

By symmetry area of quadrilateral

= 4 \times \left( {Area\,\,\Delta OAB} \right) = 4 \times {{27} \over 4} = 27\,\,

sq. units.

Q78

If the tangent at a point on the ellipse

{{{x^2}} \over {27}} + {{{y^2}} \over 3} = 1

meets the coordinate axes at A and B, and O is the origin, then the minimum area (in sq. units) of the triangle OAB is :

A

{9 \over 2}

B

3\sqrt 3

C

9\sqrt 3

D 9

Correct Answer

Option D

Solution

Equation of tangent to ellipse

{x \over {\sqrt {27} }}

cos $\theta$ +

{y \over {\sqrt 3 }}

sin $\theta$ = 1 Area bounded by line and co-ordinate axis

\Delta

=

{1 \over 2}

.

{{\sqrt {27} } \over {\cos \theta }}.{{\sqrt 3 } \over {\sin \theta }}

=

{9 \over {\sin 2\theta }}

\Delta

= will be minimum when sin 2 $\theta$ = 1

\Delta

min = 9

Q79

The eccentricity of an ellipse whose centre is at the origin is

{1 \over 2}

. If one of its directrices is x = – 4, then the equation of the normal to it at

\left( {1,{3 \over 2}} \right)

is :

A 2y – x = 2

B 4x – 2y = 1

C 4x + 2y = 7

D x + 2y = 4

Correct Answer

Option B

Solution

Given e =

{1 \over 2}

and

{a \over e}

= 4 $\therefore$

a

= 2 We have b2 =

a

2 (1 – e2) =

4\left( {1 - {1 \over 4}} \right)

= 3 $\therefore$ Equation of ellipse is

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

Now, the equation of normal at

\left( {1,{3 \over 2}} \right)

is

{{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}

$\Rightarrow$

{{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3

$\Rightarrow$ 4x – 2y = 1

Q80

If the normal to the ellipse 3x2 + 4y2 = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q(4,4) then PQ is equal to :

A

{{\sqrt {61} } \over 2}

B

{{\sqrt {221} } \over 2}

C

{{\sqrt {157} } \over 2}

D

{{5\sqrt 5 } \over 2}

Correct Answer

Option D

Solution

Equation of ellipse is

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

Normal at P(2 cos $\theta$ ,

\sqrt 3 \sin \theta

) is 2x sin $\theta$ -

\sqrt 3 y\,cos\theta

= sin $\theta$ cos $\theta$ as the normal is parallel to 2x + y = 4 $\Rightarrow$

{2 \over {\sqrt 3 }}\tan \theta = - 2

\Rightarrow \tan \theta = - \sqrt 3 \,\,......\,(i)

tangent at P(2 cos $\theta$ ,

\sqrt 3 \sin \theta

) is

\sqrt 3

x cos $\theta$ + 2y sin $\theta$ = 2

\sqrt 3

Passes through (4, 4) $\Rightarrow$ 4

\sqrt 3

cos $\theta$ + 8 sin $\theta$ = 2

\sqrt 3

....... (ii) From (i) and (ii)

\Rightarrow P\left( { - 1,{3 \over 2}} \right)\,\& \,Q(4,4)

\Rightarrow PQ = \sqrt {25 + {{25} \over 4}} = {{5\sqrt 5 } \over 2}