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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced 2018 · Paper 2 · Mathematics · Hyperbola

Let H: x ^ 2 a ^ 2 - y ^ 2 b ^ 2 = 1 , where a > b > 0 , be a...

Let H:x2a2−y2b2=1H: \dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1 , where a>b>0a > b > 0, be a hyperbola in the xy-plane whose conjugate axis LMLM subtends an angle of 60o{60}^{o} at one of its vertices NN. Let the area of the triangle LMNLMN be 434\sqrt{3}.

LIST-I LIST-II
A. The length of the conjugate axis of HH is P. 88
B. The eccentricity of HH is Q. 43\dfrac{4}{\sqrt{3}}
C. The distance between the foci of HH is R. 23\dfrac{2}{\sqrt{3}}
D. The length of the latus rectum of HH is S. 44
The correct option is:Let H: x ^ 2 a ^ 2 - y ^ 2 b ^ 2 = 1 , where a > b > 0 , be a hyperbola in the xy-plane whose conjugate axis LM

Options

  1. Aa-r;b-q;c-s;d-p;
  2. Ba-q;b-r;c-s;d-p;
  3. Ca-r;b-s;c-p;d-q;
  4. Da-s;b-r;c-p;d-q;

Correct answer

D. a-s;b-r;c-p;d-q;

Step-by-step solution

Let H: x ^ 2 a ^ 2 - y ^ 2 b ^ 2 = 1 , where a > b > 0 , be a hyperbola in the xy-plane whose conjugate axis LM
tan30o=batan {30}^{o} = \dfrac{b}{a}
⇒⇒ a=b3a = b\sqrt{3}
Now area of ∆LMN=12.2b.b3∆ LMN = \dfrac{1}{2} . 2b . b\sqrt{3}
43=3b24\sqrt{3} = \sqrt{3}{b}^{2}
⇒ b = 2 & a = 2\sqrt{3}
⇒e=1+b2a2=23⇒ e = \sqrt{1 + \dfrac{{b}^{2}}{{a}^{2}}} = \dfrac{2}{\sqrt{3}}
P. Length of conjugate axis =2b=4= 2b = 4
So P→4P \to 4
Q. Eccentricity e=23e = \dfrac{2}{\sqrt{3}}
So Q→3Q \to 3
R. Distance between foci =2ae= 2ae
=2(23)(23)=8= 2\left(2\sqrt{3}\right)\left(\dfrac{2}{\sqrt{3}}\right) = 8
So, R→1R \to 1
S. Length of latus rectum =2b2a=2(2)223=43= \dfrac{2{b}^{2}}{a} = \dfrac{2{\left(2\right)}^{2}}{2\sqrt{3}} = \dfrac{4}{\sqrt{3}}
So S→2S \to 2
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