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Mathematics · Straight lines

JEE Main 2025 — 8 April, Evening Shift — Question 25

A line passing through the point P(a,0)P(a, 0) makes an acute angle α\alpha with the positive xx-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} is

  1. Option A:

    8

  2. Option B:

    6

  3. Option C:

    5

  4. Option D:

    4

    Correct

Answer: D

Step-by-step solution

tan⁡α2=2−3\tan \frac{\alpha}{2}=2-\sqrt{3}

⇒tan⁡α=13\Rightarrow \tan \alpha=\frac{1}{\sqrt{3}}

Equation of new line: (y−0)=(2−3)(x−a)(y-0)=(2-\sqrt{3})(x-a)

y=(2−3)x−(2−3)ay=(2-\sqrt{3}) x-(2-\sqrt{3}) a

Distance from origin =12=\frac{1}{\sqrt{2}}

∣−(2−3)a4+3−43+1∣=12\left|\frac{-(2-\sqrt{3}) a}{4+3-4 \sqrt{3}+1}\right|=\frac{1}{\sqrt{2}}

∣a∣=8−432(2−3)|a|=\frac{\sqrt{8-4 \sqrt{3}}}{\sqrt{2}(2-\sqrt{3})}

∣a∣=22−32(2−3)|a|=\frac{2 \sqrt{2-\sqrt{3}}}{\sqrt{2}(2-\sqrt{3})}

∣a∣=22−3|a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}

a2=22−3=2(2+3)a^{2}=\frac{2}{2-\sqrt{3}}=2(2+\sqrt{3})

3a2tan⁡2α−23=3(4+23)×13−233 a^{2} \tan ^{2} \alpha-2 \sqrt{3}=3(4+2 \sqrt{3}) \times \frac{1}{3}-2 \sqrt{3}

=4=4

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Mathematics
Chapter
Straight lines
Topic
Various forms of Straight Line equations

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