JEE MainMathematicsStraight Lines
Let the maximum distance of any line in the family (2x - y + 1) + (ax + 2y - 4) = 0 (where is a real parameter) from the origin be 2 . Then the value of a^2 is
Options
- A4
- B64
- C100
- D36
Correct answer
D. 36
Step-by-step solution
The given family of lines is (2x - y + 1) + (ax + 2y - 4) = 0 . This family of lines passes through the fixed point P which is the intersection of the lines: 2x - y + 1 = 0 ax + 2y - 4 = 0 From the first equation, y = 2x + 1 . Substituting this into the second equation: ax + 2(2x + 1) - 4 = 0 x(a + 4) = 2 x = 2 a + 4 Then, y = 2 ( 2 a + 4 ) + 1 = a + 8 a + 4 . So, the fixed point is P ( 2 a + 4 , a + 8 a + 4 ) . The maximum distance of any line passing through a fixed point P from the origin O is the distance OP it