QJEEAdv18Ma2d39: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma2d39 Solution: Let , where a > b > 0, be a hyperbola in the XY-plane whose…

Question

Verified Solution • 2025–26 Session moderate

Let

H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

, where a > b > 0, be a hyperbola in the XY-plane whose conjugate axis LM subtends an angle of 60

^\circ

at one of its vertices N. Let the area of the

\Delta

LMN be

4\sqrt 3

.

	List - I		List - II
P.	The length of the conjugate axis of H is	1.	8
Q.	The eccentricity of H is	2.	${4 \over {\sqrt 3 }}$
R.	The distance between the foci of H is	3.	${2 \over {\sqrt 3 }}$
S.	The length of the latus rectum of H is	4.	4

Accepted Answer

We have,

Equation of hyperbola

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

It is given,

\angle LNM = 60^\circ {4 \over {\sqrt 3 }}

and Area of

\Delta LMN = 4\sqrt 3

Now,

\Delta LMN

is an equilateral triangle whose sides is 2b [

\because

\Delta LON

\cong

\Delta MOL

;

\because

\angle NLO = \angle NMO = 60^\circ

]

\therefore

Area of

\Delta LMN = {{\sqrt 3 } \over 4}{(2b)^2}

\Rightarrow

4\sqrt 3 = \sqrt 3 {b^2} \Rightarrow b = 2

Also, area of

\Delta LMN = {1 \over 2}a(2b) = ab

\Rightarrow

4\sqrt 3 = a(2)

\Rightarrow

a = 2\sqrt 3

(P) Length of conjugate axis = 2b = 2(2) = 4 (Q)

Eccentricity (e) =

\sqrt {1 + {{{b^2}} \over {{a^2}}}} = \sqrt {1 + {4 \over {12}}}

=

{4 \over {2\sqrt 3 }} = {2 \over {\sqrt 3 }}

(R) Distance between the foci

=

2ae = 2 \times 2\sqrt 3 \times {2 \over {\sqrt 3 }} = 8

(S) The length of latusrectum

=

{{2{b^2}} \over a} = {{2(4)} \over {2\sqrt 3 }} = {4 \over {\sqrt 3 }}

P

\to

4; Q

\to

3; R

\to

1; S

\to

2

Hence, option (b) is correct.

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