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The line y = mx + 4 meets the curve 3x² - (1 - 3a)xy - ay² = 0 at two points A and B such that the angle AOB = 90 degrees for all m in R except m = m1 and m = m2 (m1 < m2), where O is the origin. Which of the following statements is/are correct?

1. m1 + m2 = 10/3
2. a*m1 + m2 = 2
3. If m = 2, then area of triangle AOB = 80/7 sq. units
4. If m = 2, then area of triangle AOB = 85/7 sq. units

Correct answer: a*m1 + m2 = 2

Solution

For angle AOB = 90 deg always, the pair of lines through O (the curve) must be perpendicular: (coeff of x²) + (coeff of y²) = 0 => 3 + (-a) = 0 => a = 3. The curve becomes 3x² + 8xy - 3y² = (3x - y)(x + 3y) = 0, i.e., lines y = 3x and y = -x/3. The line y = mx + 4 is parallel to y = 3x when m = 3 (= m2) and to y = -x/3 when m = -1/3 (= m1). So m1 = -1/3, m2 = 3. Check: a*m1 + m2 = 3*(-1/3) + 3 = -1 + 3 = 2. Correct. m1 + m2 = -1/3 + 3 = 8/3 not 10/3, so option A is wrong. For m = 2: y = 2x + 4 meets y = 3x at x = 4 (A = (4,12)) and y = -x/3 at x = -12/7 (B = (-12/7, 4/7)). Area = (1/2)|OA||OB|sin(90 deg) = (1/2)*(4*sqrt(10))*(4*sqrt(10)/7) = (1/2)*(160/7) = 80/7. Option C correct.

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