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JEE Main 20254 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Hyperbola

Let the sum of the focal distances of the point P(4,3) on the...

Let the sum of the focal distances of the point P(4,3)\mathrm{P}(4,3) on the hyperbola H:x2a2−y2 b2=1\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 be 8538 \sqrt{\frac{5}{3}}. If for HH, the length of the latus rectum is ll and the product of the focal distances of the point P is m , then 9l2+6 m9 l^2+6 \mathrm{~m} is equal to :-

Options

  1. A184184
  2. B186186
  3. C185185
  4. D187187

Correct answer

C. 185185

Step-by-step solution

ex+a+ex−a=8532ex=8532e×4=853e=53b2=a2((53)2−1)b2=23a2\begin{aligned} & e x+a+e x-a=8 \sqrt{\frac{5}{3}} \\ & 2 e x=8 \sqrt{\frac{5}{3}} \\ & 2 e \times 4=8 \sqrt{\frac{5}{3}} \\ & e=\sqrt{\frac{5}{3}} \\ & b^2=a^2\left(\left(\frac{\sqrt{5}}{3}\right)^2-1\right) \\ & b^2=\frac{2}{3} a^2\end{aligned}
16a2−9 b2=1\frac{16}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=1
and b2=23a2b^2=\frac{2}{3} a^2
⇒a2=52 b2=53\Rightarrow \mathrm{a}^2=\frac{5}{2} \quad \mathrm{~b}^2=\frac{5}{3}
Now,
<br/>ℓ=2 b2aℓ2=4 b4a29ℓ2=36×259×5×29ℓ2=40 m=(ex+a)(ex−a)m=e2x2−a2=53×16−52=1456<br/>\begin{aligned}<br/>& \ell=\frac{2 \mathrm{~b}^2}{\mathrm{a}} \\ & \ell^2=\frac{4 \mathrm{~b}^4}{\mathrm{a}^2} \\ & 9 \ell^2=36 \times \frac{25}{9 \times 5} \times 2 \\ & 9 \ell^2=40 \\ & \mathrm{~m}=(\mathrm{ex}+\mathrm{a})(\mathrm{ex}-\mathrm{a}) \\ & \mathrm{m}=\mathrm{e}^2 \mathrm{x}^2-\mathrm{a}^2 \\ & =\frac{5}{3} \times 16-\frac{5}{2}=\frac{145}{6}<br/>\end{aligned}
<br/>=6 m=1459ℓ2+6 m40+145=185<br/>\begin{aligned}<br/>& =6 \mathrm{~m}=145 \\ & 9 \ell^2+6 \mathrm{~m} \\ & 40+145=185<br/>\end{aligned}
option (3)
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