Verified Solution • 2025–26 Session moderate

QJEEM24Ma1282: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1282 Solution: Let be the hyperbola, whose eccentricity is and the length of…

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 (\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

Multiple Choice Options

  • A 170
  • B 171
  • C 169
  • D 172

Step-by-Step Verified Solution

amp;H:x2a2−y2b2=−1amp;e=1+a2b2=3amp;⇒1+a2b2=3amp;⇒a2b2=2.... (1)amp;2a2b=43\begin{aligned}& H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 \\& e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3} \\& \Rightarrow 1+\frac{a^2}{b^2}=3 \\& \Rightarrow \frac{a^2}{b^2}=2 \quad \text{.... (1)}\\& \frac{2 a^2}{b}=4 \sqrt{3}\end{aligned}

Using equation (1)

amp;4b2b=43amp;⇒b=3amp;a=6amp;H:x26−y23=−1amp;α26−12=−1amp;α26=11amp;α2=66 Focus :(0,bc)(0,−bc)(0,3),(0,−3)amp;β=α2+9×α2+81amp;β=105amp;α2+β=66+105amp;=171\begin{aligned}& \frac{4 b^2}{b}=4 \sqrt{3} \\& \Rightarrow b=\sqrt{3} \\& a=\sqrt{6} \\& H: \frac{x^2}{6}-\frac{y^2}{3}=-1 \\& \frac{\alpha^2}{6}-12=-1 \\& \frac{\alpha^2}{6}=11 \\& \begin{array}{c}\alpha^2=66 \\\text { Focus }:(0, b c)(0,-b c) \\(0,3),(0,-3)\end{array} \\& \beta=\sqrt{\alpha^2+9} \times \sqrt{\alpha^2+81} \\& \beta=105 \\& \alpha^2+\beta=66+105 \\& =171\end{aligned}

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