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JEE Mathematics 2025 Question with Solution

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−233a^2 \tan^2 \alpha - 2\sqrt{3} is

  • A

    44

  • B

    55

  • C

    88

  • D

    66

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A line passes through P(a,0)P(a,0) and makes an acute angle α\alpha with the positive xx-axis. After a clockwise rotation by α2\frac{\alpha}{2} about PP, the new line has slope 2−32-\sqrt{3} and distance from the origin 12\frac{1}{\sqrt{2}}.

Find: The value of 3a2tan⁡2α−233a^2\tan^2\alpha-2\sqrt{3}.

After clockwise rotation by α2\frac{\alpha}{2}, the inclination becomes α−α2=α2\alpha-\frac{\alpha}{2}=\frac{\alpha}{2}. Hence the slope of the new line is

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

Using

tan⁡α=2tan⁡α21−tan⁡2α2\tan\alpha=\frac{2\tan\frac{\alpha}{2}}{1-\tan^2\frac{\alpha}{2}}

we get

tan⁡α=2(2−3)1−(2−3)2\tan\alpha=\frac{2(2-\sqrt{3})}{1-(2-\sqrt{3})^2}

Now,

(2−3)2=7−43(2-\sqrt{3})^2=7-4\sqrt{3}

so

tan⁡α=4−231−(7−43)=4−23−6+43=2−3−3+23\tan\alpha=\frac{4-2\sqrt{3}}{1-(7-4\sqrt{3})} =\frac{4-2\sqrt{3}}{-6+4\sqrt{3}} =\frac{2-\sqrt{3}}{-3+2\sqrt{3}}

Rationalizing,

tan⁡α=(2−3)(−3−23)(−3+23)(−3−23)=−3−3=13\tan\alpha=\frac{(2-\sqrt{3})(-3-2\sqrt{3})}{(-3+2\sqrt{3})(-3-2\sqrt{3})} =\frac{-\sqrt{3}}{-3} =\frac{1}{\sqrt{3}}

The rotated line passes through P(a,0)P(a,0) and has slope 2−32-\sqrt{3}, so its equation is

y=(2−3)(x−a)y=(2-\sqrt{3})(x-a)

or

(2−3)x−y−a(2−3)=0(2-\sqrt{3})x-y-a(2-\sqrt{3})=0

Distance of this line from the origin is

∣−(2−3)a∣(2−3)2+(−1)2=12\frac{|-(2-\sqrt{3})a|}{\sqrt{(2-\sqrt{3})^2+(-1)^2}}=\frac{1}{\sqrt{2}}

Therefore,

∣(2−3)a∣8−43=12\frac{|(2-\sqrt{3})a|}{\sqrt{8-4\sqrt{3}}}=\frac{1}{\sqrt{2}}

Squaring both sides,

(2−3)2a28−43=12\frac{(2-\sqrt{3})^2a^2}{8-4\sqrt{3}}=\frac{1}{2}

that is,

(7−43)a28−43=12\frac{(7-4\sqrt{3})a^2}{8-4\sqrt{3}}=\frac{1}{2}

So,

a2=8−432(7−43)=4−237−43a^2=\frac{8-4\sqrt{3}}{2(7-4\sqrt{3})}=\frac{4-2\sqrt{3}}{7-4\sqrt{3}}

Rationalizing,

a2=(4−23)(7+43)=4+23a^2=(4-2\sqrt{3})(7+4\sqrt{3})=4+2\sqrt{3}

Now evaluate

3a2tan⁡2α−233a^2\tan^2\alpha-2\sqrt{3}

Using a2=4+23a^2=4+2\sqrt{3} and tan⁡α=13\tan\alpha=\frac{1}{\sqrt{3}},

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

Therefore, the value of the expression is 44. The correct option is A.

Using angle interpretation directly

Given: The rotated line has slope 2−32-\sqrt{3}.

Find: The required expression without missing the geometric meaning of the slope.

Since

tan⁡15∘=2−3\tan 15^\circ=2-\sqrt{3}

the rotated line makes angle 15∘15^\circ with the positive xx-axis. But the rotated inclination is α2\frac{\alpha}{2}. Hence,

α2=15∘⇒α=30∘\frac{\alpha}{2}=15^\circ \Rightarrow \alpha=30^\circ

So,

tan⁡α=tan⁡30∘=13\tan\alpha=\tan 30^\circ=\frac{1}{\sqrt{3}}

The rotated line is

y=(2−3)(x−a)y=(2-\sqrt{3})(x-a)

Its distance from the origin is

∣a(2−3)∣1+(2−3)2=12\frac{|a(2-\sqrt{3})|}{\sqrt{1+(2-\sqrt{3})^2}}=\frac{1}{\sqrt{2}}

Using

1+(2−3)2=1+7−43=8−431+(2-\sqrt{3})^2=1+7-4\sqrt{3}=8-4\sqrt{3}

and squaring,

a2(2−3)2=12(8−43)a^2(2-\sqrt{3})^2=\frac{1}{2}(8-4\sqrt{3}) a2(7−43)=4−23a^2(7-4\sqrt{3})=4-2\sqrt{3}

Hence,

a2=4−237−43=4+23a^2=\frac{4-2\sqrt{3}}{7-4\sqrt{3}}=4+2\sqrt{3}

Finally,

3a2tan⁡2α−23=3(4+23)⋅13−23=43a^2\tan^2\alpha-2\sqrt{3}=3(4+2\sqrt{3})\cdot \frac{1}{3}-2\sqrt{3}=4

Therefore, the correct option is A.

Common mistakes

  • Assuming the new slope is tan⁡α\tan\alpha instead of tan⁡α2\tan\frac{\alpha}{2}. The line is rotated clockwise by α2\frac{\alpha}{2}, so its new inclination becomes α−α2\alpha-\frac{\alpha}{2}. Always update the angle before writing the slope.

  • Using the distance formula with the wrong constant term. For the rotated line, write it carefully as y=(2−3)(x−a)y=(2-\sqrt{3})(x-a) or (2−3)x−y−a(2−3)=0 (2-\sqrt{3})x-y-a(2-\sqrt{3})=0. Then use ∣Ax0+By0+C∣A2+B2\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} correctly.

  • Making an algebraic error while simplifying tan⁡α\tan\alpha from tan⁡α2\tan\frac{\alpha}{2}. The identity tan⁡α=2t1−t2\tan\alpha=\frac{2t}{1-t^2} with t=2−3t=2-\sqrt{3} must be handled carefully, especially while squaring and rationalizing surds.

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