A line passing through the point P(a, 0) makes an acute angle with the positive x-axis. Let this line be rotated about the point P through an angle in the clock-wise direction. If in the new position, the slope of the line is and its distance from the

Question

A line passing through the point P(a, 0) makes an acute angle $α$ with the positive x-axis. Let this line be rotated about the point P through an angle $\frac{α}{2}$ 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 $3 a^{2} \tan^{2} α - 2 \sqrt{3}$ is: [2025]

Smart Achievers · Accepted Answer

(4)

Slope of new line $= 2 - \sqrt{3}$

The new angle with x-axis is $α - \frac{α}{2} = \frac{α}{2}$

$∴ \tan (\frac{α}{2}) = 2 - \sqrt{3} = \tan (\frac{π}{12}) \Rightarrow \frac{α}{2} = \frac{π}{12} \Rightarrow α = \frac{π}{6}$

The new line passes through (a, 0) and has slope $2 - \sqrt{3}$ .

So equation of new line is

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

Now, distance of this line from origin is $\frac{1}{\sqrt{2}}$

$\Rightarrow$ $| \frac{0 + (2 - \sqrt{3}) a}{\sqrt{{(2 - \sqrt{3})}^{2} + 1^{2}}} |$ $= \frac{1}{\sqrt{2}} \Rightarrow \frac{(2 - \sqrt{3}) | a |}{\sqrt{8 - 4 \sqrt{3}}} = \frac{1}{\sqrt{2}}$

On squaring both sides, we get $\frac{{(2 - \sqrt{3})}^{2} a^{2}}{8 - 4 \sqrt{3}} = \frac{1}{2}$

$\Rightarrow 2 (7 - 4 \sqrt{3}) a^{2} = 8 - 4 \sqrt{3}$

$\Rightarrow (7 - 4 \sqrt{3}) a^{2} = 4 - 2 \sqrt{3}$

$\Rightarrow a^{2} = \frac{4 - 2 \sqrt{3}}{7 - 4 \sqrt{3}} \times \frac{7 + 4 \sqrt{3}}{7 + 4 \sqrt{3}}$

$= \frac{28 + 16 \sqrt{3} - 14 \sqrt{3} - 24}{49 - 48}$

$a^{2} = 4 + 2 \sqrt{3}$

Now, $3 a^{2} \tan^{2} α - 2 \sqrt{3} = 3 (4 + 2 \sqrt{3}) \tan^{2} \frac{π}{6} - 2 \sqrt{3}$

$= (12 + 6 \sqrt{3}) \cdot \frac{1}{3} - 2 \sqrt{3}$

$= 4 + 2 \sqrt{3} - 2 \sqrt{3} = 4$ .

Solution

1 8

2 6

3 5

4 4

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