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JEE Main 8-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 8-April-2025 Paper - Shift 24 Marks
MCQ
A line passing through the point $\mathrm{P}(\mathrm{a}, \theta)$ makes an acute angle $\alpha$ with the positive x -axis. Let this line be rotated about the point $P$ through an angle $\frac{\alpha}{2}$ in the clock-wise direction. If in the new position, the slope of the line is $2-\sqrt{3}$ and its distance from the origin is $\frac{1}{\sqrt{2}}$, then the value of $3 a^{2} \tan ^{2} \alpha-2 \sqrt{3}$ is
  • ✓
    4
  • B
    6
  • C
    5
  • D
    8
✓

Answer

Correct option: A.
4
(A) 4
Image
$m_{P R}=2-\sqrt{3}=\tan 15^{\circ}$
$\therefore \frac{\alpha}{2}=15^{\circ} \quad \Rightarrow \alpha=30^{\circ}$
equation of $P R$ :
$y=\tan 15^{\circ}(x-a)$
$y=(2-\sqrt{3})(x-a)$
$\perp$ distance from origin $=\frac{1}{\sqrt{2}}$
$\left|\frac{\sqrt{3} a-2 a}{\sqrt{4+3-4 \sqrt{3}+1}}\right|=\frac{1}{\sqrt{2}}$
$\frac{|\mathrm{a}|(2-\sqrt{3})}{2 \sqrt{(2-\sqrt{3})}}=\frac{1}{\sqrt{2}}$
$|a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}=\sqrt{2}(\sqrt{2+\sqrt{3}})$
$a^{2}=2(2+\sqrt{3})$
$3 a^{2} \tan ^{2} \alpha-2 \sqrt{3}$
$3 \times(4+2 \sqrt{3}) \cdot \frac{1}{3}-2 \sqrt{3}=4$

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