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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Straight Lines

A line passing through the point P(a, θ) makes an acute angle α...

A line passing through the point P(a,θ)\mathrm{P}(\mathrm{a}, \theta) makes an acute angle α\alpha with the positive x -axis. Let this line be rotated about the point PP through an angle α2\frac{\alpha}{2} in the clock-wise direction. If in the new position, the slope of the line is 2−32-\sqrt{3} and its distance from the origin is 12\frac{1}{\sqrt{2}}, then the value of 3a2tan⁡2α−233 a^2 \tan ^2 \alpha-2 \sqrt{3} isA line passing through the point P(a, θ) makes an acute angle α with the positive x -axis. Let this line be rotated

Options

  1. A4
  2. B6
  3. C5
  4. D8

Correct answer

A. 4

Step-by-step solution

A line passing through the point P(a, θ) makes an acute angle α with the positive x -axis. Let this line be rotated
<br/>mPR=2−3=tan⁡15∘∴α2=15∘⇒α=30∘<br/>\begin{aligned}<br/>& \mathrm{m}_{\mathrm{PR}}=2-\sqrt{3}=\tan 15^{\circ} \\ & \therefore \frac{\alpha}{2}=15^{\circ} \quad \Rightarrow \alpha=30^{\circ}<br/>\end{aligned}
equation of PR :
<br/>y=tan⁡15∘(x−a)y=(2−3)(x−a)<br/>\begin{aligned}<br/>& y=\tan 15^{\circ}(x-a) \\ & y=(2-\sqrt{3})(x-a)<br/>\end{aligned}
⊥\perp distance from origin =12=\frac{1}{\sqrt{2}}
∣3a−2a4+3−43+1∣=12∣a∣(2−3)2(2−3)=12∣a∣=22−3=2(2+3)a2=2(2+3)3a2tan⁡2α−233×(4+23)⋅13−23=4\begin{aligned} & \left|\frac{\sqrt{3} a-2 a}{\sqrt{4+3-4 \sqrt{3}+1}}\right|=\frac{1}{\sqrt{2}} \\ & \frac{|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\end{aligned}
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