Ellipse — Page 3 | JEE Mathematics

Q21

The line y = x + 1 meets the ellipse

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

at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2 is equal to :

A 20

B 12

C 11

D 8

Correct Answer

Option A

Solution

Let point (a, a + 1) as the point of intersection of line and ellipse. So,

{{{a^2}} \over 4} + {{{{(a + 1)}^2}} \over 2} = 1 \Rightarrow {a^2} + 2({a^2} + 2a + 1) = 4

\Rightarrow 3{a^2} + 4a - 2 = 0

If roots of this equation are $\alpha$ and $\beta$ . So,

P(\alpha ,\,\alpha + 1)

and

Q(\beta ,\,\beta + 1)

PQ = 4{r^2} = {(\alpha - \beta )^2} + {(\alpha - \beta )^2}

\Rightarrow 9{r^2} = {9 \over 4}(2{(\alpha - \beta )^2})

= {9 \over 2}\left[ {{{(\alpha + \beta )}^2} - 4\alpha \beta } \right]

= {9 \over 2}\left[ {{{\left( { - {4 \over 3}} \right)}^2} + {8 \over 3}} \right]

= {1 \over 2}[16 + 24] = 20

Q22

Let the maximum area of the triangle that can be inscribed in the ellipse

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

, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be

6\sqrt 3

. Then the eccentricity of the ellipse is :

A

{{\sqrt 3 } \over 2}

B

{1 \over 2}

C

{1 \over {\sqrt 2 }}

D

{{\sqrt 3 } \over 4}

Correct Answer

Option A

Solution

Given ellipse

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

$\therefore$ Let A( $\theta$ ) be the area of

\Delta

ABB' Then

A(\theta ) = {1 \over 2}4\sin \theta (a + a\cos \theta )

A'(\theta ) = a(2\cos \theta + 2{\cos ^2}\theta )

For maxima

A'(\theta ) = 0

\Rightarrow \cos \theta = 1,\,\,\cos \theta = {1 \over 2}

But for maximum area

\cos \theta = {1 \over 2}

$\therefore$

A(\theta ) = 6\sqrt 3

\Rightarrow 2{{\sqrt 3 } \over 2}\left( {a + {a \over 2}} \right) = 6\sqrt 3

\Rightarrow a = 4

$\therefore$

e = \sqrt {1 - {{{b^2}} \over {{a^2}}}} = \sqrt {1 - {4 \over {16}}} = {{\sqrt 3 } \over 2}

Q23

Let the eccentricity of the ellipse

{x^2} + {a^2}{y^2} = 25{a^2}

be b times the eccentricity of the hyperbola

{x^2} - {a^2}{y^2} = 5

, where a is the minimum distance between the curves y = ex and y = logex. Then

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

is equal to :

A

{3 \over 2}

B

{5 \over 2}

C 3

D 5

Correct Answer

Option D

Solution

Given ellipse

{x^2} + {a^2}{y^2} = 25{a^2} \Rightarrow {{{x^2}} \over {25{a^2}}} + {{{y^2}} \over {25}} = 1

eccentricity

({e_1}) = \sqrt {1 - {{{b^2}} \over {{a^2}}}}

= \sqrt {1 - {{25} \over {25{a^2}}}}

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

\Rightarrow e_1^2 = 1 - {1 \over {{a^2}}}

Also, given hyperbola,

{x^2} - {a^2}{y^2} = 5

\Rightarrow {{{x^2}} \over 5} - {{{y^2}} \over {{5 \over {{a^2}}}}} = 1

eccentricity

({e_2}) = \sqrt {1 + {{{b^2}} \over {{a^2}}}}

= \sqrt {1 + {5 \over {5{a^2}}}}

= \sqrt {1 + {1 \over {{a^2}}}}

\Rightarrow e_2^2 = 1 + {1 \over {{a^2}}}

Also given,

{e_1} = b \times {e_2}

\Rightarrow e_1^2 = {b^2} \times e_2^2

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

\Rightarrow {{{a^2} - 1} \over {{a^2}}} = {{{b^2}({a^2} + 1)} \over {{a^2}}}

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

y = \log _e^x

is inverse of

y = {e^x}

so it is mirror image of each other with respect to y = x line. Slope of tangent to y = ex curve

{{dy} \over {dx}} = {e^x}

Slope of tangent to

y = \log _e^x

curve,

{{dy} \over {dx}} = {1 \over x}

Both tangents are parallel to y = x line for minimum distance condition.

$\therefore$ Slope of y = x line = Slope of both the tangent.

$\therefore$

{{dy} \over {dx}} = {e^x} = 1 \Rightarrow {e^x} = {e^0} = x = 0

$\therefore$

y = {e^x} = {e^0} = 1

and

{{dy} \over {dx}} = {1 \over x} = 1 \Rightarrow x = 1

$\therefore$

y = \log _e^1 = 0

$\therefore$ tangent at (0, 1) point of

y = {e^x}

curve and tangent at (1, 0) point of

y = \log _e^x

curve are parallel. $\therefore$ Minimum distance between point (0, 1) and (1, 0) is

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

$\therefore$

a = \sqrt 2

So,

{b^2} = {{{a^2} - 1} \over {{a^2} + 1}} = {{2 - 1} \over {2 + 1}} = {1 \over 3}

$\therefore$

{a^2} + {1 \over {{b^2}}} = 2 + {1 \over {1/3}} = 5

Q24

If the ellipse

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

meets the line

\dfrac{x}{7}+\dfrac{y}{2 \sqrt{6}}=1

on the

x

-axis and the line

\dfrac{x}{7}-\dfrac{y}{2 \sqrt{6}}=1

on the

y

-axis, then the eccentricity of the ellipse is :

A

\dfrac{5}{7}

B

\dfrac{2 \sqrt{6}}{7}

C

\dfrac{3}{7}

D

\dfrac{2 \sqrt{5}}{7}

Correct Answer

Option A

Solution

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

meets the line

{x \over 7} + {y \over {2\sqrt 6 }} = 1

on the x-axis So,

a = 7

and

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

meets the line

{x \over 7} - {y \over {2\sqrt 6 }} = 1

on the y-axis So,

b = 2\sqrt 6

Therefore,

{e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {{24} \over {49}}

e = {5 \over 7}

Q25

The acute angle between the pair of tangents drawn to the ellipse

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

from the point

(1,3)

is :

A

\tan ^{-1}\left(\dfrac{16}{7 \sqrt{5}}\right)

B

\tan ^{-1}\left(\dfrac{24}{7 \sqrt{5}}\right)

C

\tan ^{-1}\left(\dfrac{32}{7 \sqrt{5}}\right)

D

\tan ^{-1}\left(\dfrac{3+8 \sqrt{5}}{35}\right)

Correct Answer

Option B

Solution

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

Equation of tangent having slope m.

y = mx\, \pm \,\sqrt {{5 \over 2}{m^2} + {5 \over 3}}

which passes through

(1,3)

3 = m\, \pm \sqrt {{5 \over 2}{m^2} + {5 \over 3}}

{5 \over 2}{m^2} + {5 \over 3} = 9 + {m^2} - 6m

{3 \over 2}{m^2} + 6m - {{22} \over 3} = 0

9{m^2} + 36m - 44 = 0

{m_1} + {m_2} = - 4,\,{m_1}{m_2} = - {{44} \over 9}

{({m_1} - {m_2})^2} = 16 + 4 \times {{44} \over 9} = {{320} \over 9}

Acute angle between the tangents is given by

\alpha = {\tan ^{ - 1}}\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|

= {\tan ^{ - 1}}\left| {{{{{8\sqrt 5 } \over 3}} \over {1 - {{44} \over 9}}}} \right|

= {\tan ^{ - 1}}\left( {{{24\sqrt 5 } \over {35}}} \right)

\alpha = {\tan ^{ - 1}}\left( {{{24} \over {7\sqrt 5 }}} \right)

Q26

Let a line L pass through the point of intersection of the lines

b x+10 y-8=0

and

2 x-3 y=0, \mathrm{~b} \in \mathbf{R}-\left\{\frac{4}{3}\right\}

. If the line

\mathrm{L}

also passes through the point

(1,1)

and touches the circle

17\left(x^{2}+y^{2}\right)=16

, then the eccentricity of the ellipse

\dfrac{x^{2}}{5}+\dfrac{y^{2}}{\mathrm{~b}^{2}}=1

is :

A

\dfrac{2}{\sqrt{5}}

B

\sqrt{\dfrac{3}{5}}

C

\dfrac{1}{\sqrt{5}}

D

\sqrt{\dfrac{2}{5}}

Correct Answer

Option B

Solution

{L_1}:bx + 10y - 8 = 0,\,{L_2}:2x - 3y = 0

then

L:(bx + 10y - 8) + \lambda (2x - 3y) = 0

$\because$ It passes through

(1,\,1)

$\therefore$

b + 2 - \lambda = 0 \Rightarrow \lambda = b + 2

and touches the circle

{x^2} + {y^2} = {{16} \over {17}}

\left| {{{{8^2}} \over {{{(2\lambda + b)}^2} + {{(10 - 3\lambda )}^2}}}} \right| = {{16} \over {17}}

\Rightarrow 4{\lambda ^2} + {b^2} + 4b\lambda + 100 + 9{\lambda ^2} - 60\lambda = 68

\Rightarrow 13{(b + 2)^2} + {b^2} + 4b(b + 2) - 60(b + 2) + 32 = 0

\Rightarrow 18{b^2} = 36

$\therefore$

{b^2} = 2

$\therefore$ Eccentricity of ellipse :

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

is $\therefore$

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

Q27

If the maximum distance of normal to the ellipse $$\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1, b

A

\dfrac{\sqrt{3}}{4}

B

\dfrac{1}{2}

C

\dfrac{1}{\sqrt{2}}

D

\dfrac{\sqrt{3}}{2}

Correct Answer

Option D

Solution

Equation of normal is $2 x \sec \theta-b y \operatorname{cosec} \theta=4-b^{2}$ Distance from $(0,0)=\dfrac{4-b^{2}}{\sqrt{4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta}}$ Distance is maximum if $4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta$ is minimum $\Rightarrow \tan ^{2} \theta=\dfrac{\mathrm{b}}{2}$ $\Rightarrow \dfrac{4-b^{2}}{\sqrt{4 \cdot \dfrac{b+2}{2}+b^{2} \cdot \dfrac{b+2}{b}}}=1$ $\Rightarrow 4-b^{2}=(b+2) \Rightarrow b^{2}+b-2=0$ $\Rightarrow(b+2)(b-1)=0$ $\Rightarrow b=1$ $\therefore e=\sqrt{1-\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2}$

Q28

Let the tangent and normal at the point

(3 \sqrt{3}, 1)

on the ellipse

\dfrac{x^{2}}{36}+\dfrac{y^{2}}{4}=1

meet the

y

-axis at the points

A

and

B

respectively. Let the circle

C

be drawn taking

A B

as a diameter and the line

x=2 \sqrt{5}

intersect

C

at the points

P

and

Q

. If the tangents at the points

P

and

Q

on the circle intersect at the point

(\alpha, \beta)

, then

\alpha^{2}-\beta^{2}

is equal to :

A 61

B

\dfrac{304}{5}

C 60

D

\dfrac{314}{5}

Correct Answer

Option B

Solution

\begin{aligned} & \frac{x^2}{36}+\frac{y^2}{4}=1 \\\\ & T: \frac{3 \sqrt{3} x}{36}+\frac{y}{4}=1 \\\\ & T: \frac{\sqrt{3} x}{12}+\frac{y}{4}=1 \\\\ & N: \frac{x-3 \sqrt{3}}{\frac{3 \sqrt{3}}{36}}=\frac{y-1}{\frac{1}{4}} \end{aligned}

\begin{aligned} & \frac{12 x-36 \sqrt{3}}{\sqrt{3}}=4 y-4 \\\\ & 3 x-9 \sqrt{3}=\sqrt{3} y-\sqrt{3} \\\\ & N: 3 x-\sqrt{3} y=8 \sqrt{3} \\\\ & A(0,4) \quad B(0,-8) \\\\ & \text{ C: } x^2+(y-4)(y+8)=0 \\\\ & \text{ Line } x=2 \sqrt{5} \\\\ & 20+y^2+4 y-32=0 \\\\ & y^2+4 y-12=0 \\\\ & (y+6)(y-2)=0 \end{aligned}

\begin{aligned} & P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\ & C: x^2+y^2+4 y-32=0 \\\\ & P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\ & T: x x_1+y y_1+2 y+2 y_1-32=0 \\\\ & T_1: 2 \sqrt{5} x-6 y+2 y-12-32=0 \\\\ & \quad 2 \sqrt{5} x-4 y=44 \\\\ & T_1: \sqrt{5} x-2 y=22 .......(i) \\\\ & T_2: 2 \sqrt{5} x+2 y+2 y+4-32=0 \\\\ & 2 \sqrt{5} x+4 y=28 \\\\ & T_2: \sqrt{5} x+2 y=14 .......(ii) \end{aligned}

From (i) and (ii), we get

\begin{aligned} & \alpha=\frac{18}{\sqrt{5}} \quad \beta=-2 \\\\ & \alpha^2-\beta^2=\frac{304}{5} \end{aligned}

Q29

Let

\mathrm{P}\left(\dfrac{2 \sqrt{3}}{\sqrt{7}}, \dfrac{6}{\sqrt{7}}\right), \mathrm{Q}, \mathrm{R}

and

\mathrm{S}

be four points on the ellipse

9 x^{2}+4 y^{2}=36

. Let

\mathrm{PQ}

and

\mathrm{RS}

be mutually perpendicular and pass through the origin. If

\dfrac{1}{(P Q)^{2}}+\dfrac{1}{(R S)^{2}}=\dfrac{p}{q}

, where

p

and

q

are coprime, then

p+q

is equal to :

A 143

B 147

C 137

D 157

Correct Answer

Option D

Solution

Given, points $P$ and $R$ are on the ellipse defined by $9x^2+4y^2=36$ which simplifies to $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$ .

This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis $a=3$ along the $y$ -axis and semi-minor axis $b=2$ along the $x$ -axis.

OP is the distance from origin O to point P, which is given by :

OP =r_1 = \sqrt{\left(\frac{2\sqrt{3}}{\sqrt{7}}\right)^2 + \left(\frac{6}{\sqrt{7}}\right)^2} = \sqrt{\frac{12}{7} + \frac{36}{7}} = \sqrt{\frac{48}{7}} = 2\sqrt{\frac{12}{7}}.

1.

Let's represent the given point $P\left(\dfrac{2 \sqrt{3}}{\sqrt{7}}, \dfrac{6}{\sqrt{7}}\right)$ in polar coordinates.

We can write $P$ as $(r_1 \cos \theta, r_1 \sin \theta)$ .

Since $P$ lies on the ellipse, it must satisfy the equation of the ellipse.

Substituting $x=r_1 \cos \theta$ and $y=r_1 \sin \theta$ into the equation of the ellipse gives us :

\frac{r_1^2 \cos ^2 \theta}{4}+\frac{r_1^2 \sin ^2 \theta}{9}=1

Simplifying this, we obtain :

\frac{\cos ^2 \theta}{4}+\frac{\sin ^2 \theta}{9}=\frac{7}{48} \quad \text{--- (equation 1)}

2.

Similarly, if we represent the point $R$ as $(-r_2 \sin \theta, r_2 \cos \theta)$ , (the negative sign is due to the fact that line RS is perpendicular to line PQ.

Since they are perpendicular, the angle between them is 90 degrees or $\pi/2$ radians.

In terms of sin and cos, $\sin(\theta + \pi/2) = \cos(\theta)$ and $\cos(\theta + \pi/2) = -\sin(\theta)$ .) it too should satisfy the equation of the ellipse.

We have :

\frac{r_2^2 \sin ^2 \theta}{4}+\frac{r_2^2 \cos ^2 \theta}{9}=1

Simplifying this, we obtain :

\frac{\sin ^2 \theta}{4}+\frac{\cos ^2 \theta}{9}=\frac{1}{r_2^2} \quad \text{--- (equation 2)}

3. From equations (1) and (2), we have :

\frac{1}{r_2^2}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}

4. Now, note that lines $PQ$ and $RS$ are perpendicular and pass through the origin, so $PQ = 2OP$ and $RS = 2OR$ . Thus,

\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4} \left(\frac{1}{r_1^2} + \frac{1}{r_2^2} \right)

5. Substituting the values of $r_1$ and $r_2$ , we obtain :

\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)=\frac{13}{144} = \frac{p}{q}

6. Hence, $p = 13$ and $q = 144$ . 7. So, the final answer $p+q = 13 + 144 = 157$ .

Q30

If the radius of the largest circle with centre (2,0) inscribed in the ellipse

x^2+4y^2=36

is r, then 12r

^2

is equal to :

A 72

B 92

C 115

D 69

Correct Answer

Option B

Solution

The given ellipse has the equation :

x^2+4y^2=36

We can rewrite this as :

\frac{x^2}{6^2} + \frac{y^2}{(6/2)^2} = 1

This shows that it is an ellipse centered at (0,0) with semi-major axis a = 6 along the x-axis and semi-minor axis b = 3 along the y-axis.

The equation of a circle with center (2,0) and radius r is :

(x-2)^2 + y^2 = r^2

Substituting y^2 from the ellipse equation into the circle equation gives us :

x^2 - 4x + 4 + \frac{36 - x^2}{4} = r^2

Solving this equation leads to :

3x^2 - 16x + 52 - 4r^2 = 0

For the roots of this quadratic equation to be real (which they must be, since they represent real intersection points), the discriminant (D) must be greater than or equal to zero :

D = b^2 - 4ac = (-16)^2 - 4\times3\times(52 - 4r^2) = 256 - 12\times52 + 48r^2

Setting D = 0 gives the minimum value for r (the radius of the inscribed circle) :

256 - 624 + 48r^2 = 0

48r^2 = 368

r^2 = \frac{368}{48} = \frac{23}{3}

And we're asked for the value of 12r2, so :

12r^2 = 12 \times \frac{23}{3} = 92

So, the correct answer is Option B : 92.