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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Hyperbola

Let H: (-x^2)/(a^2)+(y^2)/(b^2)=1 be the hyperbola, whose...

Let H:−x2a2+y2b2=1H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1 be the hyperbola, whose eccentricity is 3\sqrt{3} and the length of the latus rectum is 434 \sqrt{3}. Suppose the point (α,6),α>0(\alpha, 6), \alpha>0 lies on HH. If β\beta is the product of the focal distances of the point (α,6)(\alpha, 6), then α2+β\alpha^2+\beta is equal to

Options

  1. A172
  2. B171
  3. C169
  4. D170

Correct answer

B. 171

Step-by-step solution

H:y2b2−x2a2=1,e=3e=1+a2b2=3⇒a2b2=2a2=2b2 length of L.R. =2a2b=43a=6\begin{aligned} & \mathrm{H}: \frac{\mathrm{y}^2}{\mathrm{ b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1, \mathrm{e}=\sqrt{3} \\ & \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{ b}^2}}=\sqrt{3} \Rightarrow \frac{\mathrm{a}^2}{\mathrm{ b}^2}=2 \\ & \qquad \mathrm{a}^2=2 \mathrm{ b}^2 \\ & \text{ length of L.R. }=\frac{2 \mathrm{a}^2}{\mathrm{ b}}=4 \sqrt{3} \\ & \qquad a=\sqrt{6}\end{aligned}P(\alpha, 6)lieonlie on\frac{y^2}{3}-\frac{x^2}{6}=1 12−α26=1⇒α2=66 Foci =(0,± be )=(0,3)(0,−3)\begin{aligned} & 12-\frac{\alpha^2}{6}=1 \Rightarrow \alpha^2=66 \\ & \text{ Foci }=(0, \pm \text{ be })=(0,3) &(0,-3) \end{aligned}LetLetd_1 \& d_2befocaldistancesofbe focal distances of\mathrm{P}(\alpha, 6) d1=α2+(6+be)2,d2=α2+(6−be)2d1=66+81,d2=66+9β=d1d2=147×75=105α2+β=66+105=171\begin{aligned} & \mathrm{d}_1=\sqrt{\alpha^2+(6+b e)^2}, \mathrm{ d}_2=\sqrt{\alpha^2+(6-b e)^2} \\ & \mathrm{ d}_1=\sqrt{66+81}, \mathrm{ d}_2=\sqrt{66+9} \\ & \beta=\mathrm{d}_1 \mathrm{ d}_2=\sqrt{147 \times 75}=105 \\ & \alpha^2+\beta=66+105=171\end{aligned}$SECTION - B
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