JEE Advanced 2018 · Paper 2 · Mathematics Q17

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MathematicsList match+3 / -1Question 53 of 54

Let $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , 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 triangle $LMN$ be $4\sqrt{3}$ .

The correct option is:

LIST-I

P
The length of the conjugate axis of $H$ is
Q
The eccentricity of $H$ is
R
The distance between the foci of $H$ is
S
The length of the latus rectum of $H$ is

LIST-II

1
8
2
$\frac{4}{\sqrt{3}}$
3
$\frac{2}{\sqrt{3}}$
4
4

P → 4; Q → 2; R → 1; S → 3

P → 4; Q → 3; R → 1; S → 2

P → 4; Q → 1; R → 3; S → 2

P → 3; Q → 4; R → 2; S → 1

Answer and worked solution

Official answer: B

Find $a$ and $b$ . $L=(0,b)$ , $M=(0,-b)$ and $N=(a,0)$ . The $60^\circ$ angle at $N$ is split equally by the $x$ -axis, so $\tan30^\circ=\dfrac ba$ and $a=\sqrt3b$ .

Area. $\tfrac12(2b)(a)=\sqrt3b^2=4\sqrt3$ , so $b=2$ and $a=2\sqrt3$ .

P. The conjugate axis is $2b=4$ : 4.
Q. $e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ : 3.
R. The distance between the foci is $2ae=2(2\sqrt3)\dfrac2{\sqrt3}=8$ : 1.
S. The latus rectum is $\dfrac{2b^2}a=\dfrac8{2\sqrt3}=\dfrac4{\sqrt3}$ : 2.

(B)

Topics:Hyperbola

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JEE Advanced 2018 · Paper 2, Mathematics question 17