Ellipse — Page 9 | JEE Mathematics

Q81

The tangent and normal to the ellipse 3x2 + 5y2 = 32 at the point P(2, 2) meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is :

A

{{14} \over 3}

B

{{16} \over 3}

C

{{68} \over {15}}

D

{{34} \over {15}}

Correct Answer

Option C

Solution

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

6x + 10yy' = 0

\Rightarrow y' = {{ - 3x} \over {5y}}

\Rightarrow y{'_{(2,2)}} = - {3 \over 5}

Tangent

(y - 2) = - {3 \over 5}(x - 2) \Rightarrow Q\left( {{{16} \over 3},0} \right)

Normal

(y - 2) = {5 \over 3}(x - 2) \Rightarrow R\left( {{{4} \over 5},0} \right)

$\therefore$ Area =

{1 \over 2}(QR) \times 2 = QR = {{68} \over {15}}

Q82

If the line x – 2y = 12 is tangent to the ellipse

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

at the point

\left( {3, - {9 \over 2}} \right)

, then the length of the latus rectum of the ellipse is :

A 5

B 9

C

8\sqrt 3

D

12\sqrt 2

Correct Answer

Option B

Solution

Equation of tangent at

\left( {3, - {9 \over 2}} \right)

to

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

is

{{3x} \over {{a^2}}} - {{{y^9}} \over {2{b^2}}} = 1

which is equivalent to x – 2y = 12

{3 \over {{a^2}}} = {{ - 9} \over {2{b^2}( - 2)}} = {1 \over {12}}

(On comparing)

{a^2} = 3 \times 12

and

{b^2} = {{9 \times 12} \over 4}

$\Rightarrow$ a = 6 and b =

3\sqrt 3

So latus rectum =

{{2{b^2}} \over a} = {{2 \times 27} \over 6} = 9

Q83

If the tangent to the parabola y2 = x at a point (

\alpha

,

\beta

), (

\beta

> 0) is also a tangent to the ellipse, x2 + 2y2 = 1, then

\alpha

is equal to :

A

\sqrt 2 + 1

B

\sqrt 2 - 1

C

2\sqrt 2 + 1

D

2\sqrt 2 - 1

Correct Answer

Option A

Solution

Point P( $\alpha$ , $\beta$ ) is on the parabola y2 = x $\therefore$

{\beta ^2} = \alpha

...........(1) Equation of tangent to the parabola y2 = x at ( $\alpha$ , $\beta$ ) is T = 0

\beta y = {{x + \alpha } \over 2}

$\Rightarrow$

2\beta y = x + \alpha

$\Rightarrow$

y = {1 \over {2\beta }}x + {\alpha \over {2\beta }}

For this straight line, m =

{1 \over {2\beta }}

and c =

{\alpha \over {2\beta }}

This tangent is also a tangent to ellipse x2 + 2y2 = 1. $\Rightarrow$

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

$\therefore$

{a^2} = 1

and

{b^2} = {1 \over 2}

. We know the condition of tangency on ellipse is

{c^2} = {a^2}{m^2} + {b^2}

$\Rightarrow$

{\left( {{\alpha \over {2\beta }}} \right)^2} = 1{\left( {{1 \over {2\beta }}} \right)^2} + {1 \over 2}

$\Rightarrow$

{{{\alpha ^2}} \over {4{\beta ^2}}} = {1 \over {4{\beta ^2}}} + {1 \over 2}

$\Rightarrow$

{\alpha ^2} = 1 + 2{\beta ^2}

$\Rightarrow$

{\alpha ^2} = 1 + 2\alpha

$\Rightarrow$

{\alpha ^2} - 2\alpha - 1 = 0

$\Rightarrow$

\alpha = 1 + \sqrt 2

and

\alpha = 1 - \sqrt 2

Q84

If the tangents on the ellipse 4x2 + y2 = 8 at the points (1, 2) and (a, b) are perpendicular to each other, then a2 is equal to :

A

{{2} \over {17}}

B

{{64} \over {17}}

C

{{128} \over {17}}

D

{{4} \over {17}}

Correct Answer

Option A

Solution

Given, Equation of ellipse 4x2 + y2 = 8 We know equation of tangent at any point (x1, y1) is 4xx1 + yy1 = 8 $\therefore$ Equation of tangent at point (1, 2) is 4x + 2y = 8 $\Rightarrow$ 2x + y = 4 $\therefore$ Slope of this tangent = -2 Equation of tangent at point (a, b) is 4ax + by = 8 Slope of this tangent =

- {{4a} \over b}

As tangent at (1, 2) and (a, b) are perpendicular $\therefore$

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

$\Rightarrow$ b = -8

a

As point (a, b) lies on the ellipse $\therefore$

4{a^2} + {b^2} = 8

$\Rightarrow$

4{a^2} + 64{a^2} = 8

$\Rightarrow$

{a^2} = {8 \over {68}}

=

{2 \over {17}}

Q85

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0,5

\sqrt 3

), then the length of its latus rectum is :

A 5

B 8

C 10

D 6

Correct Answer

Option A

Solution

Focus (0, be) = (0, 5

\sqrt 3

) $\therefore$ be = 5

\sqrt 3

$\Rightarrow$ b2e2 = 75 As here b > a so e2 =

1 - {{{a^2}} \over {{b^2}}}

$\therefore$ b2

\left( {1 - {{{a^2}} \over {{b^2}}}} \right)

= 75 $\Rightarrow$ b2 - a2 = 75 .......(

1) Given that, difference of the lengths of major axis and minor axis is 10. $\therefore$ 2b - 2a = 10 $\Rightarrow$ b - a = 5 .......(

2) From (1), (b + a)(b - a) = 75 $\Rightarrow$ (b + a)5 = 75 $\Rightarrow$ (b + a) = 15 .......(

3) From (2) and (3) we get, b = 10, a = 5 $\therefore$ Length of its latus rectum =

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

=

{{2 \times 25} \over {10}}

= 5

Q86

If 3x + 4y = 12

\sqrt 2

is a tangent to the ellipse

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

for some

a

\in

R, then the distance between the foci of the ellipse is :

A

2\sqrt 5

B

2\sqrt 7

C 4

D

2\sqrt 2

Correct Answer

Option B

Solution

3x + 4y = 12

\sqrt 2

$\Rightarrow$ y =

- {{3x} \over 4} + 3\sqrt 2

is tangent to

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

$\therefore$ c2 = a2m2 + b2 $\Rightarrow$

{\left( {3\sqrt 2 } \right)^2} = {a^2}{\left( { - {3 \over 4}} \right)^2} + 9

$\Rightarrow$ a2 = 16 Also e =

\sqrt {1 - {{{b^2}} \over {{a^2}}}}

=

\sqrt {1 - {9 \over {16}}}

=

{{\sqrt 7 } \over 4}

Distance between focii = 2ae =

2 \times 4 \times {{\sqrt 7 } \over 4}

=

2\sqrt 7

Q87

Let the line y = mx and the ellipse 2x2 + y2 = 1 intersect at a ponit P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at

\left( { - {1 \over {3\sqrt 2 }},0} \right)

and (0,

\beta

), then

\beta

is equal to :

A

{{\sqrt 2 } \over 3}

B

{2 \over 3}

C

{{2\sqrt 2 } \over 3}

D

{2 \over {\sqrt 3 }}

Correct Answer

Option A

Solution

Let P be (x1 , y1) Equation of normal at P is

{x \over {2{x_1}}} - {y \over {{y_1}}} = {1 \over 2} - 1

It passes through

\left( { - {1 \over {3\sqrt 2 }},0} \right)

$\therefore$

{{ - 1} \over {6\sqrt 2 {x_1}}} = - {1 \over 2}

$\Rightarrow$ x1 =

{1 \over {3\sqrt 2 }}

Also using 2

x_1^2

+

y_1^2

= 1 $\Rightarrow$

y_1^2

= 1 -

{2 \over {18}}

$\Rightarrow$ y1 =

{{2\sqrt 2 } \over 3}

(0, $\beta$ ) is on the normal. So 0 -

{\beta \over {{{2\sqrt 2 } \over 3}}}

=

- {1 \over 2}

$\Rightarrow$ $\beta$ =

{{\sqrt 2 } \over 3}

Q88

The length of the minor axis (along y-axis) of an ellipse in the standard form is

{4 \over {\sqrt 3 }}

. If this ellipse touches the line, x + 6y = 8; then its eccentricity is :

A

{1 \over 3}\sqrt {{{11} \over 3}}

B

{1 \over 2}\sqrt {{5 \over 3}}

C

\sqrt {{5 \over 6}}

D

{1 \over 2}\sqrt {{{11} \over 3}}

Correct Answer

Option D

Solution

Let the equation of ellipse

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

, (

a > b

) Given 2b =

{4 \over {\sqrt 3 }}

$\Rightarrow$ b =

{2 \over {\sqrt 3 }}

We know, Equation of tangent y = mx $\pm$

\sqrt {{a^2}{m^2} + {b^2}}

....(1) Given tangent is x + 6y = 8 $\Rightarrow$ y =

- {1 \over 6}x + {8 \over 6}

.....(2) By comparing (1) and (2), m =

- {1 \over 6}

and

{{a^2}{m^2} + {b^2}}

=

{{16} \over 9}

$\Rightarrow$

{{a^2}\left( {{1 \over {36}}} \right) + {4 \over 3}}

=

{{16} \over 9}

$\Rightarrow$

{{a^2} = 16}

$\therefore$ e =

\sqrt {1 - {{{b^2}} \over {{a^2}}}}

=

\sqrt {1 - {{{4 \over 3}} \over {16}}}

=

\sqrt {{{11} \over {12}}}

=

{1 \over 2}\sqrt {{{11} \over 3}}

Q89

Let

f(x)=x^2+9, g(x)=\dfrac{x}{x-9}

and

\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3)

. If

\mathrm{e}

and

l

denote the eccentricity and the length of the latus rectum of the ellipse

\dfrac{x^2}{\mathrm{a}}+\dfrac{y^2}{\mathrm{~b}}=1

, then

8 \mathrm{e}^2+l^2

is equal to.

A 6

B 12

C 8

D 16

Correct Answer

Option C

Solution

\begin{aligned} & g(10)=10 \\ & a=f(g(10))=f(10)=109 \\ & f(3)=18 \\ & b=g(f(3))=g(18)=2 \\ & \frac{x^2}{109}+\frac{y^2}{2}=1 \\ & e=\sqrt{1-\frac{2}{109}}=\sqrt{\frac{107}{109}} \\ & I=\frac{2 b^2}{a}=\frac{2 \times 2}{\sqrt{109}} \end{aligned}

\begin{aligned} 8 e^2+l^2 & =\frac{8 \times 107}{109}+\frac{16}{109} \\ & =8 \end{aligned}

Q90

Let

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

(a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function,

\phi \left( t \right) = {5 \over {12}} + t - {t^2}

, then a2 + b2 is equal to :

A 145

B 126

C 135

D 116

Correct Answer

Option B

Solution

Given ellipse

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

(a > b) Length of latus rectum

= {{2{b^2}} \over a} = 10

\phi (t) = {5 \over {12}} + t - {t^2}

= {8 \over {12}} - {\left( {t - {1 \over 2}} \right)^2}

$\therefore$

\phi {(t)_{\max }} = {8 \over {12}} = {2 \over 3}

$\therefore$ eccentricity (e) =

{2 \over 3}

Also,

{e^2} = 1 - {{{b^2}} \over {{a^2}}}

\Rightarrow {4 \over 9} = 1 - {{{b^2}} \over {{a^2}}}

\Rightarrow {{{b^2}} \over {{a^2}}} = {5 \over 9}

\Rightarrow {{{b^2}} \over a} \times {1 \over a} = {5 \over 9}

\Rightarrow {5 \over a} = {5 \over 9}

\Rightarrow a = 9

$\therefore$

{b^2} = 5 \times 9 = 45

$\therefore$

{a^2} + {b^2} = 81 + 45 = 126