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

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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]

1 8

2 6

3 5

4 4

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

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

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

Want Us to Email you About Special Offers & Updates?