Let the length of a latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be 10. If its eccentricity is the minimum value of the function $f(t) = t^2 + t + \frac{11}{12}$, $t \in \mathbb{

Given that the length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is 10, we have: $ \frac{2b^2}{a} = 10 \quad \Rightarrow \quad 5a = b^2 \tag{1} $ Next, consider the function $ f(t) = t^2 + t + \frac{11}{12} $. To find its minimum value, we calculate the derivative: $ \frac{df(t)}{dt} = 2t + 1 = 0 \quad \Rightarrow \quad t = \frac{-1}{2} $ Plugging $t = -\frac{1}{2}$ into $f(t)$ gives the minimum value: $ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \

Let for two distinct values of p the lines $y=x+\mathrm{p}$ touch the ellipse $\mathrm{E}: \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$ at the points A and B . Let the line $y=x$ intersect E at the points C and

$E: \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$ $$\begin{aligned} & T: y=m x \pm \sqrt{16 m^2+9} \\ & y=x+p \\ & \Rightarrow m=1 \\ & \Rightarrow p= \pm \sqrt{16+9} \\ & = \pm 5 \end{aligned}$$ $T: y=x \pm 5$ will to cut the $E$ at $A\left(-\frac{16}{5}, \frac{9}{5}\right)$ $$B\left(\frac{16}{5},-\frac{9}{5}\right)$$ Also, $y=x$ will cut the $E$ at $C\left(\frac{12}{5}, \frac{12}{5}\right)$ $$D\left(-\frac{12}{5},-\frac{12}{5}\right)$$ $A B C D$ in not give in cyclic order $\therefore

Ellipse — Page 6 | JEE Mathematics

Q51

Let the length of a latus rectum of an ellipse

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

be 10. If its eccentricity is the minimum value of the function

f(t) = t^2 + t + \dfrac{11}{12}

,

t \in \mathbb{R}

, then

a^2 + b^2

is equal to :

A 115

B 120

C 125

D 126

Correct Answer

Option D

Solution

Given that the length of the latus rectum of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is 10, we have: $\dfrac{2b^2}{a} = 10 \quad \Rightarrow \quad 5a = b^2 \tag{1}$ Next, consider the function $f(t) = t^2 + t + \dfrac{11}{12}$ .

To find its minimum value, we calculate the derivative: $\dfrac{df(t)}{dt} = 2t + 1 = 0 \quad \Rightarrow \quad t = \dfrac{-1}{2}$ Plugging $t = -\dfrac{1}{2}$ into $f(t)$ gives the minimum value: $f\left(-\dfrac{1}{2}\right) = \left(-\dfrac{1}{2}\right)^2 + \left(-\dfrac{1}{2}\right) + \dfrac{11}{12} = \dfrac{1}{4} - \dfrac{1}{2} + \dfrac{11}{12} = \dfrac{3 - 6 + 11}{12} = \dfrac{8}{12} = \dfrac{2}{3}$ Thus, the eccentricity $e$ of the ellipse is $\dfrac{2}{3}$ , so $e^2 = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}$ .

Using the eccentricity formula for an ellipse: $e^2 = \dfrac{1 - b^2}{a^2} \quad \Rightarrow \quad \dfrac{4}{9} = \dfrac{1 - b^2}{a^2}$ Rearranging gives: $b^2 = a^2 \left(1 - \dfrac{4}{9}\right) = a^2 \cdot \dfrac{5}{9}$ From equation $(1)$ , $b^2 = 5a$ .

Substituting, we have: $5a = a^2 \cdot \dfrac{5}{9} \quad \Rightarrow \quad a^2 = 9a$ Solving for $a$ , $a = 9, \quad \text{and therefore } b^2 = 5a = 45 \quad \Rightarrow \quad b = \sqrt{45} = 3\sqrt{5}$ Finally, calculate $a^2 + b^2$ : $a^2 + b^2 = 81 + 45 = 126$

Q52

Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentricity of the ellipse

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

is :

A

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

B

\dfrac{1}{2}

C

\dfrac{\sqrt{7}}{4}

D

\dfrac{3}{4}

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Total trangles }=\Rightarrow={ }^{\mathrm{h}} \mathrm{C}_3 \\ & \text{ Total auadrilaterals }={ }^{\mathrm{h}} \mathrm{C}_4=\mathrm{q} \\ & { }^{\mathrm{n}} \mathrm{C}_3+{ }^{\mathrm{n}} \mathrm{C}_4=126 \Rightarrow{ }^{\mathrm{n}+1} \mathrm{C}_4=126 \\ & \Rightarrow \mathrm{n}+1=9 \Rightarrow \mathrm{n}=8 \\ & \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{n}}=1 \Rightarrow \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{8}=1 \\ & \mathrm{e}=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}} \end{aligned}

Q53

A line passing through the point

P(\sqrt{5}, \sqrt{5})

intersects the ellipse

\dfrac{x^2}{36}+\dfrac{y^2}{25}=1

at

A

and

B

such that

(P A) \cdot(P B)

is maximum. Then

5\left(P A^2+P B^2\right)

is equal to :

A 290

B 377

C 338

D 218

Correct Answer

Option C

Solution

\begin{aligned} & \frac{x-\sqrt{5}}{\cos \theta}=\frac{y-\sqrt{5}}{\sin \theta}=r \\ & x=r \cos \theta+\sqrt{5} \\ & y=r \sin \theta+\sqrt{5} \\ & 25 x^2+36 y^2=900,(x, y) \text{ lie on } E \\ & r^2\left[25 \cos ^2 \theta+36 \sin ^2 \theta\right]+r[50 \sqrt{5} \cos \theta+72 \sqrt{5} \sin \theta] \\ & \quad+25[5]+36[5]=900 \\ & r_1 r_2=\frac{900-305}{25+11 \sin ^2 \theta},\left(r_1 r_2\right)_{\max }=\frac{595}{25}=\frac{119}{5} \\ & \left|r_1 \cdot r_2\right|_{\max }=\left(\frac{595}{25}\right)=\left(\frac{119}{5}\right) \text{ at } \theta=0 \end{aligned}

\begin{aligned} & \frac{x^2}{36}+\frac{y^2}{25}=1 \\ & \text{ At } y=\sqrt{5} \\ & \frac{x^2}{36}+\frac{1}{5}=1 \\ & \Rightarrow \frac{x^2}{36}=\frac{4}{5} \Rightarrow x= \pm \frac{12}{\sqrt{5}} \\ & P A^2+P B^2=(P A+P B)^2-2 P A \cdot P B=\left(\frac{24}{\sqrt{5}}\right)^2-2 \cdot \frac{119}{5} \\ & 5\left(P A^2+P B^2\right)=5\left(\frac{24^2}{5}-\frac{2.119}{5}\right)=24^2-238 \\ & =338 . \end{aligned}

Q54

Let for two distinct values of p the lines

y=x+\mathrm{p}

touch the ellipse

\mathrm{E}: \dfrac{x^2}{4^2}+\dfrac{y^2}{3^2}=1

at the points A and B . Let the line

y=x

intersect E at the points C and D . Then the area of the quadrilateral

A B C D

is equal to :

A 48

B 20

C 24

D 36

Correct Answer

Option C

Solution

$E: \dfrac{x^2}{4^2}+\dfrac{y^2}{3^2}=1$

\begin{aligned} & T: y=m x \pm \sqrt{16 m^2+9} \\ & y=x+p \\ & \Rightarrow m=1 \\ & \Rightarrow p= \pm \sqrt{16+9} \\ & = \pm 5 \end{aligned}

$T: y=x \pm 5$ will to cut the $E$ at $A\left(-\dfrac{16}{5}, \dfrac{9}{5}\right)$

B\left(\frac{16}{5},-\frac{9}{5}\right)

Also, $y=x$ will cut the $E$ at $C\left(\dfrac{12}{5}, \dfrac{12}{5}\right)$

D\left(-\frac{12}{5},-\frac{12}{5}\right)

$A B C D$ in not give in cyclic order $\therefore$ it does not form any quadrilateral $\therefore \quad$ No option should match If order is not considered then Area $=24$ sq. unit.

Q55

The centre of a circle C is at the centre of the ellipse

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

. Let C pass through the foci

F_1

and

F_2

of E such that the circle

C

and the ellipse

E

intersect at four points. Let P be one of these four points. If the area of the triangle

\mathrm{PF}_1 \mathrm{~F}_2

is 30 and the length of the major axis of

E

is 17 , then the distance between the foci of

E

is :

A 12

B 26

C 13

D

\dfrac{13}{2}

Correct Answer

Option C

Solution

\begin{aligned} & x^2+\frac{a^2 y^2}{b^2}=a^2 \\ & \Rightarrow y^2\left(1-\frac{a^2}{b^2}\right)=a^2\left(e^2-1\right)=a^2\left(1-\frac{b^2}{a^2}-1\right) \\ & =-b^2 \\ & \Rightarrow \frac{y^2\left(b^2-a^2\right)}{b^2}=-b^2 \Rightarrow y^2=\frac{b^4}{\left(a^2-b^2\right)} \\ & \text{ Height }=|y|=\frac{b^2}{\sqrt{a^2-b^2}} \end{aligned}

\begin{aligned} &\begin{aligned} & \text{ Area }=(2 a e) \times \frac{1}{2} \times \frac{b^2}{\sqrt{a^2-b^2}}=30 \\ & =\frac{a b^2 e}{a \sqrt{1-\frac{b^2}{a^2}}}=b^2, a=\frac{17}{2} \end{aligned}\\ &\text{ Distance between foci }=2 a e\\ &=17 \sqrt{1-\frac{b^2}{a^2}}=17 \sqrt{1-\frac{30 \times 4}{289}}=13 \end{aligned}

Q56

The length of the latus-rectum of the ellipse, whose foci are

(2,5)

and

(2,-3)

and eccentricity is

\dfrac{4}{5}

, is

A

\dfrac{50}{3}

B

\dfrac{18}{5}

C

\dfrac{6}{5}

D

\dfrac{10}{3}

Correct Answer

Option B

Solution

To find the length of the latus rectum of the ellipse, we first recognize that the foci of the ellipse are given as $F_1:(2,5)$ and $F_2:(2,-3)$ .

This indicates that the major axis is aligned along the $y$ -axis.

Calculate the distance between the foci: $F_1F_2 = 8$ The formula involving the distance between the foci and the eccentricity is: $F_1F_2 = 2be$ Here, the eccentricity $e$ is given as $\dfrac{4}{5}$ .

Thus, we can solve for $b$ : $8 = 2b \cdot \dfrac{4}{5} \implies b = \dfrac{8}{2 \times \dfrac{4}{5}} = 5$ Determine $a^2$ : Using the relationship between eccentricity, semi-minor axis $b$ , and semi-major axis $a$ : $e^2 = 1 - \dfrac{a^2}{b^2} = 1 - \dfrac{a^2}{25} = \dfrac{16}{25}$ Solving for $a^2$ : $1 - \dfrac{a^2}{25} = \dfrac{16}{25} \implies \dfrac{a^2}{25} = \dfrac{9}{25} \implies a^2 = 9$ Thus, $a = 3$ .

Compute the length of the latus rectum: The formula for the length of the latus rectum $L$ is: $L = \dfrac{2a^2}{b}$ Substituting the known values: $L = \dfrac{2 \times (9)}{5} = \dfrac{18}{5}$ Therefore, the length of the latus rectum of the ellipse is $\dfrac{18}{5}$ .

Q57

Let

C

be the circle of minimum area enclosing the ellipse

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

with eccentricity

\dfrac{1}{2}

and foci

( \pm 2,0)

. Let

P Q R

be a variable triangle, whose vertex

P

is on the circle

C

and the side

Q R

of length

2 a

is parallel to the major axis of

E

and contains the point of intersection of

E

with the negative

y

-axis. Then the maximum area of the triangle

P Q R

is :

A

8(3+\sqrt{2})

B

8(2+\sqrt{3})

C

6(3+\sqrt{2})

D

6(2+\sqrt{3})

Correct Answer

Option B

Solution

\begin{aligned} &\text{ Area }=\frac{1}{2} \times(a+b) \cdot 2 a=a(a+b)\\ &\text{ Since, } e=\frac{1}{2} a e=2 \Rightarrow a=4, b=2 \sqrt{3}\\ &\Rightarrow \text{ Area }=4(4+2 \sqrt{3})=8(2+\sqrt{3}) \end{aligned}

Q58

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is :

A

\sqrt 3

B

3\sqrt 2

C

{3 \over {\sqrt 2 }}

D

2\sqrt 3

Correct Answer

Option B

Solution

Distance between foci = 2ae = 6 $\Rightarrow$ ae = 3 .....(1) Distance between directrices =

{{2a} \over e}

= 12 $\Rightarrow$

{a \over e}

= 6 .....(2) from (1) and (2) a2 = 18 also a2e2 = 9 $\Rightarrow$ 18e2 = 9 $\Rightarrow$ e2 =

{1 \over 2}

We know e2 = 1 -

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

$\therefore$

{1 \over 2}

= 1 -

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

$\Rightarrow$ b2 = 9 $\therefore$ Length of latus rectum =

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

=

{{2 \times 9} \over {\sqrt {18} }}

=

3\sqrt 2

Q59

An ellipse, with foci at (0, 2) and (0, –2) and minor axis of length 4, passes through which of the following points?

A

\left( {2,\sqrt 2 } \right)

B

\left( {2,2\sqrt 2 } \right)

C

\left( {\sqrt 2 ,2} \right)

D

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

Correct Answer

Option C

Solution

Let

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

is the equation of ellipse, focii

(0, \pm 2)

Given 2a = 4 $\Rightarrow$ a = 2 e2 = 1 -

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

$\Rightarrow$ b2e2 = b2 - a2 4 = b2 - 4 b2 = 8 $\because$ equation of ellipse is

{{{x^2}} \over 4} + {{{y^2}} \over 8} = 1

then it passes through

\left( {\sqrt 2 ,2} \right)

Q60

If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

A

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

B

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

C

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

D

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

Correct Answer

Option B

Solution

Equation of general tangent on ellipse

{x \over {a\,\sec \theta }} + {y \over {b\cos ec\theta }} = 1

a = \sqrt 2 ,\,\,b = 1

\Rightarrow {x \over {\sqrt 2 \sec \theta }} + {y \over {\cos ec\theta }} = 1

Let the midpoint be (h, k)

h = {{\sqrt 2 \sec \theta } \over 2} \Rightarrow \cos \theta = {1 \over {\sqrt 2 h}}

and

k = {{\cos ec\theta } \over 2} \Rightarrow \sin \theta = {1 \over {2k}}

$\because$

{\sin ^2}\theta + {\cos ^2}\theta = 1

$\Rightarrow$

{1 \over {2{h^2}}} + {1 \over {4{k^2}}} = 1

$\Rightarrow$

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