Question

A line passing through the point $ P(a, 0) $ 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 $ 3a^2 \tan^2 \alpha - 2\sqrt{3} $ is

• A. 4
• B. 5
• C. 8
• D. 6

Hint:

- When rotating lines, remember angle addition formulas - Distance from point to line formula: \( \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \) - For \( \tan 15^\circ \), exact value is \( 2 - \sqrt{3} \) - Simplify radicals by rationalizing denominators

Answer 1

The Correct Option is A

1. Understand the Geometry and Transformations

• We start with a line passing through $(a, 0)$ with an angle $\alpha$ to the positive x-axis.
• It's rotated clockwise by $\alpha/2$ around $(a, 0)$.
• The new slope is given, and the distance of the new line from the origin is given.

2. Find the Initial Slope ($\tan \alpha$)

• Let the slope of the rotated line be $m$. We are given $m = 2 - \sqrt{3}$. This is equal to $\tan(\alpha - \alpha/2) = \tan(\alpha/2)$.
• Therefore, $\tan(\alpha/2) = 2 - \sqrt{3}$.
• Using the identity $\tan(\alpha) = \frac{2\tan(\alpha/2)}{1 - \tan^2(\alpha/2)}$: \[ \tan(\alpha) = \frac{2(2 - \sqrt{3})}{1 - (2 - \sqrt{3})^2} = \frac{4 - 2\sqrt{3}}{1 - (4 - 4\sqrt{3} + 3)} = \frac{4 - 2\sqrt{3}}{1 - 7 + 4\sqrt{3}} = \frac{4 - 2\sqrt{3}}{-6 + 4\sqrt{3}} = \frac{2 - \sqrt{3}}{-3 + 2\sqrt{3}} \] Rationalize the denominator by multiplying both numerator and denominator by $(-3 - 2\sqrt{3})$: \[ \tan(\alpha) = \frac{(2 - \sqrt{3})(-3 - 2\sqrt{3})}{(-3 + 2\sqrt{3})(-3 - 2\sqrt{3})} = \frac{-6 - 4\sqrt{3} + 3\sqrt{3} + 6}{9 - 12} = \frac{-\sqrt{3}}{-3} = \frac{1}{\sqrt{3}} \]
• Since $\alpha$ is acute, $\alpha = \frac{\pi}{6}$.

3. Find the Equation of the Rotated Line

• The slope of the rotated line is $m = 2 - \sqrt{3}$.
• The equation of the line passing through $(a, 0)$ with slope $m$ is $y = m(x - a)$ or $y = (2 - \sqrt{3})(x - a)$ or $(2-\sqrt{3})x - y - a(2-\sqrt{3}) = 0$.
• The distance of this line from the origin $(0, 0)$ is given as $\frac{1}{\sqrt{2}}$. Use the distance formula: \[ \frac{|A*0 + B*0 + C|}{\sqrt{A^2 + B^2}} = \frac{1}{\sqrt{2}} \] \[ \frac{|-(2-\sqrt{3})a|}{\sqrt{(2-\sqrt{3})^2 + (-1)^2}} = \frac{1}{\sqrt{2}} \] \[ \frac{|(2-\sqrt{3})a|}{\sqrt{4 - 4\sqrt{3} + 3 + 1}} = \frac{1}{\sqrt{2}} \] \[ \frac{|(2-\sqrt{3})a|}{\sqrt{8 - 4\sqrt{3}}} = \frac{1}{\sqrt{2}} \] \[ \frac{(2-\sqrt{3})^2 a^2}{8 - 4\sqrt{3}} = \frac{1}{2} \] \[ \frac{(7 - 4\sqrt{3})a^2}{8 - 4\sqrt{3}} = \frac{1}{2} \] \[ a^2 = \frac{8 - 4\sqrt{3}}{2(7 - 4\sqrt{3})} \] \[ a^2 = \frac{4 - 2\sqrt{3}}{7 - 4\sqrt{3}} \] Rationalize the denominator by multiplying both numerator and denominator by $(7 + 4\sqrt{3})$: \[ a^2 = \frac{(4 - 2\sqrt{3})(7 + 4\sqrt{3})}{49 - 48} = 28 + 16\sqrt{3} - 14\sqrt{3} - 24 = 4 + 2\sqrt{3} \]

4. Evaluate the Expression

• We need to find the value of $3a^2 \tan^2(\alpha) - 2\sqrt{3}$.
• We know $a^2 = 4 + 2\sqrt{3}$ and $\tan^2(\alpha) = \tan^2(\frac{\pi}{6}) = (\frac{1}{\sqrt{3}})^2 = \frac{1}{3}$.
• Substituting these values: \[ 3(4 + 2\sqrt{3})\left(\frac{1}{3}\right) - 2\sqrt{3} = (4 + 2\sqrt{3}) - 2\sqrt{3} = 4 \]

Answer: The value of $3a^2 \tan^2 \alpha - 2\sqrt{3}$ is 4.
So the answer is option 1.