Ellipse — Page 4 | JEE Mathematics

Q31

Consider ellipses

\mathrm{E}_{k}: k x^{2}+k^{2} y^{2}=1, k=1,2, \ldots, 20

. Let

\mathrm{C}_{k}

be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse

\mathrm{E}_{k}

. If

r_{k}

is the radius of the circle

\mathrm{C}_{k}

, then the value of

\sum\limits_{k=1}^{20} \dfrac{1}{r_{k}^{2}}

is :

A 2870

B 3210

C 3320

D 3080

Correct Answer

Option D

Solution

We have, $E_K=K x^2+K^2 y^2=1, K=1,2, \ldots 20$ $\Rightarrow \dfrac{x^2}{\dfrac{1}{K}}+\dfrac{y^2}{\dfrac{1}{K^2}}=1$ Equation of $A B$ is $\dfrac{x}{\dfrac{1}{\sqrt{K}}}+\dfrac{y}{\dfrac{1}{K}}=1$ or $\sqrt{K} x+K y=1$ $r_K$ is the radius of circle $C_K$ , So, $r_K=$ perpendicular distance from $(0,0)$ to $\mathrm{AB}$

\begin{aligned} r_K & =\frac{|0+0-1|}{\sqrt{(\sqrt{K})^2+K^2}} \\\\ & =\frac{1}{\sqrt{K+K^2}} \\\\ \frac{1}{r_K^2} & =K+K^2 \end{aligned}

\begin{aligned} \sum_{K=1}^{20} \frac{1}{r_K^2} & =\sum_{K=1}^{20} K+\sum_{K=1}^{20} K^2 \\\\ & =\frac{20 \times 21}{2}+\frac{20 \times 21 \times 41}{6} \\\\ & =210+2870=3080 \end{aligned}

Q32

Let a circle of radius 4 be concentric to the ellipse

15 x^{2}+19 y^{2}=285

. Then the common tangents are inclined to the minor axis of the ellipse at the angle :

A

\dfrac{\pi}{4}

B

\dfrac{\pi}{3}

C

\dfrac{\pi}{6}

D

\dfrac{\pi}{12}

Correct Answer

Option B

Solution

We have, equation of ellipse : $15 x^2+19 y^2=285$ or $\dfrac{x^2}{19}+\dfrac{y^2}{15}=1$ Let the coordinate of center of circle be $(0,0)$ .

Equation of circle is $x^2+y^2=16$ Equation of tangent of ellipse is

\begin{gathered} y=m x \pm \sqrt{19 m^2+15} \text{ or } \\\\ m x-y \pm \sqrt{19 m^2+15}=0 \end{gathered}

It is also tangent to the circle $x^2+y^2=16$ Perpendicular distance from center of circle to tangent $=4$

\frac{\left|0-0 \pm \sqrt{19 m^2+15}\right|}{\sqrt{m^2+1}}=4

On squaring both side, we get

\begin{gathered} 19 m^2+15=16 m^2+16 \\\\ 3 m^2=1 \\\\ m^2=\frac{1}{3} \\\\ m= \pm \frac{1}{\sqrt{3}} \end{gathered}

\tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6} \text{ with } X \text{-axis or }

$\dfrac{\pi}{3}$ with $Y$ -axis Hence, the common tangents are inclined to the minor axis of the ellipse at an angle of $\dfrac{\pi}{3}$ .

Q33

Let the ellipse

E:{x^2} + 9{y^2} = 9

intersect the positive x and y-axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is

{m \over n}

, where m and n are coprime, then

m - n

is equal to :

A 15

B 16

C 17

D 18

Correct Answer

Option C

Solution

The given equation of the ellipse is

\begin{aligned} & x^2+9 y^2=9 ~..........(i)\\\\ & \Rightarrow \frac{x^2}{9}+\frac{y^2}{1}=1 \end{aligned}

Now, equation of line $A B$ is

x+3 y=3 ~

...........(ii) Now, point of intersection of line $x+3 y=3$ and circle

\begin{array}{lr} &x^2+y^2=9 \\\\ &\therefore (3-3 y)^2+y^2=9 \\\\ &\Rightarrow 9-18 y+9 y^2+y^2=9 \\\\ &\Rightarrow -18 y+10 y^2=0 \\\\ &\Rightarrow 18 y=10 y^2 \\\\ &\Rightarrow y=0, \frac{9}{5} \end{array}

Now, from figure, we clearly see that vertices $A, P$ and $O$ makes a $\triangle A P O$ , whose area is $\dfrac{1}{2} \times$ Base $\times$ Height

\begin{aligned} & \text{ Area }=\frac{1}{2} \times 3 \times \frac{9}{5}=\frac{27}{10}=\frac{m}{n} ~~~~[Given]\\\\ & \therefore m=27, n=10 \\\\ & \Rightarrow m-n=27-10=17 \end{aligned}

Q34

Let the line

2 x+3 y-\mathrm{k}=0, \mathrm{k}>0

, intersect the

x

-axis and

y

-axis at the points

\mathrm{A}

and

\mathrm{B}

, respectively. If the equation of the circle having the line segment

A B

as a diameter is

x^2+y^2-3 x-2 y=0

and the length of the latus rectum of the ellipse

x^2+9 y^2=k^2

is

\dfrac{m}{n}

, where

m

and

n

are coprime, then

2 \mathrm{~m}+\mathrm{n}

is equal to

A 12

B 13

C 11

D 10

Correct Answer

Option C

Solution

Equation of circle with

A B

as diameter

\begin{aligned} & \left(x-\frac{k}{2}\right) x+y\left(y-\frac{k}{3}\right)=0 \\ & \Rightarrow x^2+y^2-\frac{k x}{2}-\frac{k y}{3}=0 \end{aligned}

Comparing,

k=6

Latus rectum of ellipse

\begin{aligned} & x^2+9 y^2=k^2=6^2 \\ & \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1 \\ & \text{ L.R }=\frac{2 b^2}{a}=\frac{2 \times 4}{6}=\frac{4}{3} \\ & m=4 \\ & n=3 \\ & 2 m+n=8+3=11 \end{aligned}

Q35

In a group of 100 persons 75 speak English and 40 speak Hindi. Each person speaks at least one of the two languages. If the number of persons, who speak only English is

\alpha

and the number of persons who speak only Hindi is

\beta

, then the eccentricity of the ellipse

25\left(\beta^{2} x^{2}+\alpha^{2} y^{2}\right)=\alpha^{2} \beta^{2}

is :

A

\dfrac{\sqrt{129}}{12}

B

\dfrac{3 \sqrt{15}}{12}

C

\dfrac{\sqrt{119}}{12}

D

\dfrac{\sqrt{117}}{12}

Correct Answer

Option C

Solution

Let $E$ be the person speak, English $\therefore n(E)=75$ and $H$ be the person speak Hindi $\therefore n(H)=40$ Let number of persons who speak both English and Hindi are $t$ .

$\begin{array}{rlrl} &\therefore \alpha+t+\beta =100 ........(i) \\\\ & \alpha+t =75........(i) \\\\ &\text{ and }\beta+t =40 ........(iii)\end{array}$ From Equations (i) and (ii), $\beta=25$ From Equations (i) and (iii), $\alpha=60$ and from Eq. (i), $t=15$ We have, equation of ellipse

\begin{aligned} 25\left(\beta^2 x^2+\alpha^2 y^2\right) & =\alpha^2 \beta^2 \\\\ \Rightarrow 25\left(\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}\right) & =1 \end{aligned}

$\begin{aligned} & \Rightarrow 25\left(\dfrac{x^2}{3600}+\dfrac{y^2}{625}\right)=1 \\\\ & \Rightarrow \dfrac{x^2}{144}+\dfrac{y^2}{25}=1 \\\\ & \therefore \text{ Eccentricity, } e=\sqrt{1-\dfrac{b^2}{a^2}}=\sqrt{1-\dfrac{25}{144}}=\dfrac{\sqrt{119}}{12}\end{aligned}$

Q36

Let

\mathrm{P}

be a point on the ellipse

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

. Let the line passing through

\mathrm{P}

and parallel to

y

-axis meet the circle

x^2+y^2=9

at point

\mathrm{Q}

such that

\mathrm{P}

and

\mathrm{Q}

are on the same side of the

x

-axis. Then, the eccentricity of the locus of the point

R

on

P Q

such that

P R: R Q=4: 3

as

P

moves on the ellipse, is :

A

\dfrac{13}{21}

B

\dfrac{\sqrt{139}}{23}

C

\dfrac{\sqrt{13}}{7}

D

\dfrac{11}{19}

Correct Answer

Option C

Solution

$\begin{aligned} & \mathrm{h}=3 \cos \theta \\\\ & \mathrm{k}=\dfrac{18}{7} \sin \theta\end{aligned}$ $\begin{aligned} & \therefore \text{ locus }=\dfrac{\mathrm{x}^2}{9}+\dfrac{49 \mathrm{y}^2}{324}=1 \\\\ & \mathrm{e}=\sqrt{1-\dfrac{324}{49 \times 9}}=\dfrac{\sqrt{117}}{21}=\dfrac{\sqrt{13}}{7}\end{aligned}$

Q37

Let

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

be an ellipse, whose eccentricity is

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

and the length of the latusrectum is

\sqrt{14}

. Then the square of the eccentricity of

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

is :

A 3

B

{7 \over 2}

C

{3 \over 2}

D

{5 \over 2}

Correct Answer

Option C

Solution

Given the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a > b$ , the eccentricity $e$ is given by the formula: $e = \sqrt{1 - \left(\dfrac{b}{a}\right)^2}$ It is provided that the eccentricity $e$ is $\dfrac{1}{\sqrt{2}}$ (given), so we can equate the two expressions for eccentricity: $\dfrac{1}{\sqrt{2}} = \sqrt{1 - \left(\dfrac{b}{a}\right)^2}$ Squaring both sides to eliminate the square root gives: $\dfrac{1}{2} = 1 - \left(\dfrac{b}{a}\right)^2$ $\left(\dfrac{b}{a}\right)^2 = 1 - \dfrac{1}{2}$ $\left(\dfrac{b}{a}\right)^2 = \dfrac{1}{2}$ Taking the square root on both sides: $\dfrac{b}{a} = \dfrac{1}{\sqrt{2}}$ $a = b\sqrt{2}$ Now, for the ellipse, the length of the latus rectum is given by the formula: $\text{Length of Latus Rectum (L)} = \dfrac{2b^2}{a}$ It's provided that the length of the latus rectum $L$ is $\sqrt{14}$ , so substitute the known values to find $b$ : $\sqrt{14} = \dfrac{2b^2}{b\sqrt{2}} = \dfrac{2b}{\sqrt{2}}$ $b\sqrt{2} = \sqrt{14}$ $b^2 = \dfrac{14}{2}$ $b^2 = 7$ And since $a = b\sqrt{2}$ , we can find $a^2$ : $a^2 = (b\sqrt{2})^2$ $a^2 = 7 \cdot 2$ $a^2 = 14$ Now we have an ellipse with $a^2 = 14$ and $b^2 = 7$ .

The equation of a hyperbola similar to the given ellipse but with the terms subtracted is: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ For the hyperbola, the square of the eccentricity $e'$ is given by: $(e')^2 = 1 + \dfrac{b^2}{a^2}$ Substitute the values we've found for $a^2$ and $b^2$ into the formula for the square of the hyperbola's eccentricity: $(e')^2 = 1 + \dfrac{b^2}{a^2}$ $(e')^2 = 1 + \dfrac{7}{14}$ $(e')^2 = 1 + \dfrac{1}{2}$ $(e')^2 = \dfrac{3}{2}$ Therefore, the square of the eccentricity of the hyperbola is $\dfrac{3}{2}$ , which corresponds to option C.

Q38

The length of the chord of the ellipse

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

, whose mid point is

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

, is equal to :

A

\dfrac{\sqrt{1691}}{5}

B

\dfrac{\sqrt{2009}}{5}

C

\dfrac{\sqrt{1541}}{5}

D

\dfrac{\sqrt{1741}}{5}

Correct Answer

Option A

Solution

Equation of chord with given middle point.

\begin{aligned} & T=S_1 \\ & \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \\ & \frac{8 x+5 y}{200}=\frac{8+2}{200} \\ & y=\frac{10-8 x}{5} \quad \text{.... (i)} \end{aligned}

\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1

(put in original equation)

\begin{aligned} & \frac{16 x^2+100+64 x^2-160 x}{400}=1 \\ & 4 x^2-8 x-15=0 \\ & x=\frac{8 \pm \sqrt{304}}{8} \\ & x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8} \end{aligned}

Similarly,

y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}

\begin{aligned} & \mathrm{y}_1=\frac{2-\sqrt{304}}{5} ; \mathrm{y}_2=\frac{2+\sqrt{304}}{5} \\ & \text{ Distance }=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2} \\ & =\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5} \\ \end{aligned}

Q39

Let

P

be a parabola with vertex

(2,3)

and directrix

2 x+y=6

. Let an ellipse

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

, of eccentricity

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

pass through the focus of the parabola

P

. Then, the square of the length of the latus rectum of

E

, is

A

\dfrac{512}{25}

B

\dfrac{656}{25}

C

\dfrac{385}{8}

D

\dfrac{347}{8}

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Slope of axis }=\frac{1}{2} \\ & y-3=\frac{1}{2}(x-2) \\ & \Rightarrow 2 y-6=x-2 \\ & \Rightarrow 2 y-x-4=0 \\ & 2 x+y-6=0 \\ & 4 x+2 y-12=0 \\ & \alpha+1.6=4 \Rightarrow \alpha=2.4 \\ & \beta+2.8=6 \Rightarrow \beta=3.2 \end{aligned}

Ellipse passes through

(2.4,3.2)

\Rightarrow \frac{\left(\frac{24}{10}\right)^2}{\mathrm{a}^2}+\frac{\left(\frac{32}{10}\right)^2}{\mathrm{~b}^2}=1

..... (1) Also

1-\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2}=\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2}

\Rightarrow a^2=2 b^2

Put in (1)

\Rightarrow b^2=\frac{328}{25}

\Rightarrow\left(\frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)^2=\frac{4 \mathrm{~b}^2}{\mathrm{a}^2} \times \mathrm{b}^2=4 \times \frac{1}{2} \times \frac{328}{25}=\frac{656}{25}

Q40

Let

A(\alpha, 0)

and

B(0, \beta)

be the points on the line

5 x+7 y=50

. Let the point

P

divide the line segment

A B

internally in the ratio

7:3

. Let

3 x-25=0

be a directrix of the ellipse

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

and the corresponding focus be

S

. If from

S

, the perpendicular on the

x

-axis passes through

P

, then the length of the latus rectum of

E

is equal to,

A

\dfrac{25}{3}

B

\dfrac{25}{9}

C

\dfrac{32}{5}

D

\dfrac{32}{9}

Correct Answer

Option C

Solution

\left.\begin{array}{l} \mathrm{A}=(10,0) \\ \mathrm{B}=\left(0, \frac{50}{7}\right) \end{array}\right\} \mathrm{P}=(3,5)

\begin{aligned} & \text{ ae }=3 \\ & \frac{\mathrm{a}}{\mathrm{e}}=\frac{25}{3} \\ & \mathrm{a}=5 \\ & \mathrm{~b}=4 \end{aligned}

Length of

L R=\frac{2 b^2}{a}=\frac{32}{5}