Ellipse — Page 5 | JEE Mathematics

Q41

If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

A

\dfrac{1}{\sqrt{3}}

B

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

C

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

D

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

Correct Answer

Option B

Solution

\begin{aligned} & 2 b=a e \\ & \frac{b}{a}=\frac{e}{2} \\ & e=\sqrt{1-\frac{e^2}{4}} \\ & e=\frac{2}{\sqrt{5}} \end{aligned}

Q42

If

\alpha x+\beta y=109

is the equation of the chord of the ellipse

\dfrac{x^2}{9}+\dfrac{y^2}{4}=1

, whose mid point is

\left(\dfrac{5}{2}, \dfrac{1}{2}\right)

. then

\alpha+\beta

is equal to :

A 37

B 46

C 72

D 58

Correct Answer

Option D

Solution

\begin{aligned} &\text{ Equation of chord } \mathrm{T}=\mathrm{S}_1\\ &\begin{aligned} & \frac{5}{2}\left(\frac{\mathrm{x}}{9}\right)+\frac{1}{2}\left(\frac{\mathrm{y}}{4}\right)=\frac{25}{36}+\frac{1}{16} \\ & \Rightarrow \frac{5 \mathrm{x}}{18}+\frac{\mathrm{y}}{8}=\frac{100+9}{144}=\frac{109}{144} \\ & \Rightarrow 40 \mathrm{x}+18 \mathrm{y}=109 \\ & \Rightarrow \alpha=40, \beta=18 \\ & \Rightarrow \alpha+\beta=58 \end{aligned} \end{aligned}

Q43

Let

\mathrm{E}: \dfrac{x^2}{\mathrm{a}^2}+\dfrac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}

and

\mathrm{H}: \dfrac{x^2}{\mathrm{~A}^2}-\dfrac{y^2}{\mathrm{~B}^2}=1

. Let the distance between the foci of E and the foci of

H

be

2 \sqrt{3}

. If

a-A=2

, and the ratio of the eccentricities of

E

and

H

is

\dfrac{1}{3}

, then the sum of the lengths of their latus rectums is equal to :

A 10

B 7

C 9

D 8

Correct Answer

Option D

Solution

We are given an ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a>b,

and a hyperbola

\frac{x^2}{A^2}-\frac{y^2}{B^2}=1.

It is stated that “the distance between the foci of $E$ and the foci of $H$ is $2\sqrt{3}$ .”

A natural interpretation is that the foci of the ellipse are separated by

2\sqrt{a^2-b^2},

and those of the hyperbola by

2\sqrt{A^2+B^2},

and each distance is equal to $2\sqrt{3}$ .

That is, we have $2\sqrt{a^2-b^2}=2\sqrt{3}\quad \Longrightarrow\quad a^2-b^2=3,$ and $2\sqrt{A^2+B^2}=2\sqrt{3}\quad \Longrightarrow\quad A^2+B^2=3.$ We are also given that

a-A=2,

and that the ratio of the eccentricities is

\frac{e_E}{e_H}=\frac{1}{3},

where the eccentricity of the ellipse is

e_E=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{3}}{a},

and the eccentricity of the hyperbola is

e_H=\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{3}}{A}.

Thus the ratio becomes

\frac{e_E}{e_H}=\frac{\sqrt{3}/a}{\sqrt{3}/A}=\frac{A}{a}=\frac{1}{3}.

This implies

a=3A.

Now, using the condition $a-A=2$ together with $a=3A$ , we get

3A-A=2 \quad\Longrightarrow\quad 2A=2 \quad\Longrightarrow\quad A=1.

Thus,

a=3.

Next, for the ellipse we have

a^2-b^2=3 \quad \Longrightarrow\quad 9-b^2=3 \quad\Longrightarrow\quad b^2=6.

For the hyperbola,

A^2+B^2=3 \quad \Longrightarrow\quad 1+B^2=3 \quad\Longrightarrow\quad B^2=2.

The length of the latus rectum is given by the following formulas: For the ellipse:

L_E=\frac{2b^2}{a},

For the hyperbola:

L_H=\frac{2B^2}{A}.

Substitute the computed values: For the ellipse:

L_E=\frac{2\times6}{3}=\frac{12}{3}=4,

For the hyperbola:

L_H=\frac{2\times2}{1}=4.

The sum of the lengths of the latus rectums is then

L_E+L_H=4+4=8.

Thus, the answer is

8.

Q44

If the midpoint of a chord of the ellipse

\dfrac{x^2}{9}+\dfrac{y^2}{4}=1

is

(\sqrt{2}, 4 / 3)

, and the length of the chord is

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

, then

\alpha

is :

A 26

B 18

C 22

D 20

Correct Answer

Option C

Solution

\begin{aligned} &\text{ If } \mathrm{m}\left(\sqrt{2}, \frac{4}{3}\right) \text{ than equation of of } \mathrm{AB} \text{ is }\\ &\begin{aligned} & \mathrm{T}=\mathrm{S}_1 \\ & \frac{\mathrm{x} \sqrt{2}}{9}+\frac{\mathrm{y}}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^2}{9}+\frac{\left(\frac{4}{3}\right)^2}{4} \\ & \frac{\sqrt{2} \mathrm{x}}{9}+\frac{\mathrm{y}}{3}=\frac{2}{9}+\frac{4}{9} \end{aligned} \end{aligned}

\begin{aligned} &\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3} \text{ put in ellipse }\\ &\begin{aligned} & \text{ So, } \frac{x^2}{9}+\frac{(6-\sqrt{2} x)^2}{9 \times 4}=1 \\ & 4 x^2+36+2 x^2-12 \sqrt{2} x=36 \\ & 6 x^2-12 \sqrt{2} x=0 \\ & 6 x(x-2 \sqrt{2})=0 \\ & x=0 \& x=2 \sqrt{2} \\ & \text{ So } y=2 \quad y=\frac{2}{3} \\ & \text{ Length of chord }=\sqrt{(2 \sqrt{2}-0)^2+\left(\frac{2}{3}-2\right)^2} \\ & \quad=\sqrt{8+\frac{16}{9}} \\ & \quad=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22} \text{ so } \alpha=22 \end{aligned} \end{aligned}

Q45

The length of the chord of the ellipse

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

, whose mid-point is

\left(1, \dfrac{1}{2}\right)

, is :

A

\dfrac{2}{3} \sqrt{15}

B

\dfrac{1}{3} \sqrt{15}

C

\sqrt{15}

D

\dfrac{5}{3} \sqrt{15}

Correct Answer

Option A

Solution

\begin{aligned} &\begin{aligned} & \mathrm{T}=\mathrm{S}_1 \\ & \frac{\mathrm{x} \cdot 1}{4}+\frac{\mathrm{y} \cdot 1 / 2}{2}=\frac{1}{4}+\frac{1}{8} \\ & \mathrm{x}+\mathrm{y}=\frac{3}{2} \end{aligned}\\ &\text{ solve with ellipse }\\ &\begin{aligned} \mathrm{PR} & =\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2} \\ & =\sqrt{2}\left|\mathrm{x}_2-\mathrm{x}_1\right| \end{aligned} \end{aligned}

\begin{aligned} & y_2=\frac{3}{2}-x_2 \\ & y_1=\frac{3}{2}-x_1 \\ & y_2-y_1=x_2-x_1 \\ & x^2+2 y^2=4 \\ & x^2+2\left(\frac{3}{2}-x\right)^2=4 \\ & 6 x^2-12 x+1=0 \\ & x_1+x_2=2 \\ & x_1 x_2=1 / 6 \\ & \left|x_2-x_1\right|=\sqrt{\left(x_2+x_1\right)^2-4 x_1 x_2} \\ & \quad=\sqrt{4-4 / 6} \\ & P R=\sqrt{2} \cdot 2 \cdot \frac{\sqrt{5}}{\sqrt{2} \sqrt{3}}=\frac{2}{3} \sqrt{15} \end{aligned}

Q46

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :

A

\dfrac{3}{\sqrt{19}}

B

\dfrac{\sqrt{3}}{16}

C

\dfrac{4}{\sqrt{17}}

D

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

Correct Answer

Option C

Solution

The length of the minor axis is equal to one-fourth of the distance between the foci.

Mathematically, this can be expressed as: $2b = \dfrac{1}{4}(2ae)$ This simplifies to: $b = \dfrac{ae}{4}$ Given that the relationship between $b$ , $a$ , and $e$ is: $\dfrac{b^2}{a^2} = 1 - e^2$ Substitute $b = \dfrac{ae}{4}$ into the equation: $\left(\dfrac{ae}{4}\right)^2 = a^2 \cdot \left(1 - e^2\right)$ Expanding and simplifying gives: $\dfrac{a^2 e^2}{16} = a^2(1 - e^2)$ Divide both sides by $a^2$ : $\dfrac{e^2}{16} = 1 - e^2$ Rearrange and solve for $e^2$ : $\dfrac{e^2}{16} + e^2 = 1$ $\dfrac{17e^2}{16} = 1$ Solve for $e$ : $e^2 = \dfrac{16}{17}$ $e = \dfrac{4}{\sqrt{17}}$ Thus, the eccentricity of the ellipse is: $\dfrac{4}{\sqrt{17}}$

Q47

Let the product of the focal distances of the point

\left(\sqrt{3}, \dfrac{1}{2}\right)

on the ellipse

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,(a>b)

, be

\dfrac{7}{4}

. Then the absolute difference of the eccentricities of two such ellipses is

A

\dfrac{1-2 \sqrt{2}}{\sqrt{3}}

B

\dfrac{1-\sqrt{3}}{\sqrt{2}}

C

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

D

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

Correct Answer

Option C

Solution

\begin{aligned} & \text{ Product of focal distances }=\left(\mathrm{a}+\mathrm{ex}_1\right)\left(\mathrm{a}-\mathrm{ex}_1\right) \\ & =\mathrm{a}^2-\mathrm{e}^2 \mathrm{x}_1^2=\mathrm{a}^2-\mathrm{e}^2(3) \\ & =\mathrm{a}^2-3 \mathrm{e}^2=\frac{7}{4} \Rightarrow \mathrm{a}^2=\frac{7}{4}+3 \mathrm{e}^2 \\ & \Rightarrow 4 \mathrm{a}^2=7+12 \mathrm{e}^2 \\ & \&\left(\sqrt{3}, \frac{1}{2}\right) \text{ lines on } \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \\ & \therefore \frac{3}{\mathrm{a}^2}+\frac{1}{4 \mathrm{~b}^2}=1 \\ & \frac{3}{\mathrm{a}^2}+\frac{1}{4\left(\mathrm{a}^2\right)\left(1-\mathrm{e}^2\right)}=1 \\ & 12\left(1-\mathrm{e}^2\right)+1=4 \mathrm{a}^2\left(1-\mathrm{e}^2\right) \\ & 13-12 \mathrm{e}^2=\left(7+12 \mathrm{e}^2\right)\left(1-\mathrm{e}^2\right) \\ & \Rightarrow 13-12 \mathrm{e}^2=7-7 \mathrm{e}^2+12 \mathrm{e}^2-12 \mathrm{e}^4 \\ & \Rightarrow 12 \mathrm{e}^4-17 \mathrm{e}^2+6=0 \\ & \therefore \mathrm{e}^2=\frac{17 \pm \sqrt{289-288}}{24}=\frac{17 \pm 1}{24}=\frac{3}{4} \& \frac{2}{3} \\ & \therefore \mathrm{e}=\frac{\sqrt{3}}{2} \& \sqrt{\frac{2}{3}} \\ & \therefore \text{ difference }==\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2 \sqrt{2}}{2 \sqrt{3}} \end{aligned}

Q48

The equation of the chord, of the ellipse

\dfrac{x^2}{25}+\dfrac{y^2}{16}=1

, whose mid-point is

(3,1)

is :

A

5 x+16 y=31

B

48 x+25 y=169

C

4 x+122 y=134

D

25 x+101 y=176

Correct Answer

Option B

Solution

\begin{aligned} &\text{ Equation of chord with given middle point }\\ &\begin{aligned} & \mathrm{T}=\mathrm{S}_1 \\ & \Rightarrow \frac{3 \mathrm{x}}{25}+\frac{\mathrm{y}}{16}-1=\frac{9}{25}+\frac{1}{16}-1 \\ & 48 \mathrm{x}+25 \mathrm{y}=144+25 \\ & 48 \mathrm{x}+25 \mathrm{y}=169 \text{ Ans. } \end{aligned} \end{aligned}

Q49

Let the ellipse

3x^2 + py^2 = 4

pass through the centre

C

of the circle

x^2 + y^2 - 2x - 4y - 11 = 0

of radius

r

. Let

f_1, f_2

be the focal distances of the point

C

on the ellipse. Then

6f_1f_2 - r

is equal to

A 78

B 68

C 70

D 74

Correct Answer

Option C

Solution

\begin{aligned} &E: \frac{x^2}{4 / 3}+\frac{y^2}{4 / P}=1\\ &\text{ Centre of circle }(1,2) \text{, radius }\\ &\mathrm{r}=\sqrt{1+4+11} \end{aligned}

\mathrm{r}=4

$\because$ E pass from centre $(1,2)$

\therefore \frac{3}{4}+\mathrm{P}=1

$\mathrm{P}=\dfrac{1}{4} \quad \therefore$ vertical ellipse

\begin{aligned} & \mathrm{e}=\sqrt{1-\frac{4 / 3}{16}}=\sqrt{1-\frac{1}{12}}=\sqrt{\frac{11}{12}} \\ & \therefore \text{ Focal distance of } \mathrm{C}(\mathrm{~h}, \mathrm{k}) \\ & =\mathrm{b} \pm \mathrm{ek} \\ & \mathrm{~F}_1=4+\sqrt{\frac{11}{12}} \times 2 \\ & \mathrm{~F}_2=4-\sqrt{\frac{11}{12}} \times 2 \\ & \therefore \mathrm{~F}_1 \mathrm{~F}_2=16-\frac{11}{3}=\frac{37}{3} \end{aligned}

$\therefore 6 \mathrm{~F}_1 \mathrm{~F}_2-\mathrm{r}=74-4=70$

Q50

If

S

and

S^{\prime}

are the foci of the ellipse

\dfrac{x^2}{18}+\dfrac{y^2}{9}=1

and P be a point on the ellipse, then

\min \left(S P \cdot S^{\prime} P\right)+\max \left(S P \cdot S^{\prime} P\right)

is equal to :

A

3(6+\sqrt{2})

B

3(1+\sqrt{2})

C 27

D 9

Correct Answer

Option C

Solution

\begin{aligned} & a=3 \sqrt{2}, b=3 \\ & \Rightarrow \quad e=\frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} & P S \cdot P S^{\prime}=2 a=6 \sqrt{2} \\ & \frac{P S+P S^{\prime}}{2} \geq \sqrt{P S \cdot P S^{\prime}} \\ & \Rightarrow\left(P S \times P S^{\prime}\right) \max =18 \end{aligned}

Minima happens when $P$ lies on major axis

\begin{aligned} & \Rightarrow \quad P=(3 \sqrt{2}, 0) \\ & P S=(3 \sqrt{2}-3) \cdot P S^{\prime}=(3 \sqrt{2}+3) \\ & \left(P S \cdot P S^{\prime}\right) \min =9 \\ & \left(P S \cdot P S^{\prime}\right) \min +\left(P S \cdot P S^{\prime}\right) \max =27 \end{aligned}

Option (1)