Q62
jee_main_2025_08_april_evening
Rotation of Coordinate Axes / Distance from Origin
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 mathrmP through an angle fracalpha2 in the clock-wise direction. If in the new position, the slope of the line is 2 - sqrt3 and its distance from the origin is frac1sqrt2, then the value of 3mathrma^2tan^2alpha - 2sqrt3 is
Solution
### Related Formula
textDistance from origin d = frac|C|sqrtA^2+B^2
### Core Logic
Decode slope angle fields to define vector shifts. Match variables to isolate focal root segments accurately as provided in the reference layout steps.
### Step 1: Identify Phase Shift Angles
Given final tracking position slope component is:
m = 2 - sqrt3 = tan 15^circ
Because clockwise translation turns angular coordinates down by half increments:
alpha - fracalpha2 = 15^circ implies alpha = 30^circ
### Step 2: Construct Rotated Vector Path Line
The line equation through point (a,0) with final angle scaling is:
y = (2-sqrt3)(x-a) implies (2-sqrt3)x - y - a(2-sqrt3) = 0
### Step 3: Solve Origin Geometric Profiles
Impose perpendicular bounds tracking distance rule:
left| fracsqrt3a - 2asqrt4+3-4sqrt3+1 right| = frac1sqrt2 implies a^2 = 2(2+sqrt3)
Target formulation output text string solution:
3a^2 tan^2 30^circ - 2sqrt3 = 3 times 2(2+sqrt3) times frac13 - 2sqrt3 = 4
{{SOL_IMG_62}}
### Pattern Recognition
Recognizing standard trigonometric slopes (\tan 15^\circ = 2-\sqrt3) bypasses computational bottlenecks instantly when working with line updates.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
