Question

Let straight line y = mx + 4 meets the curve 3x2 – (1 – 3a) xy – ay2 = 0 at two points A and B such that where m1 < m2 and ‘O’ is the origin. find m1+m2 and area of triangle AOB if m=2

Answer 1

m_1+m_2 = 20/3, Area = 16/5

Answer 2

1. Identify the given curve as a pair of straight lines through the origin and find the sum and product of their slopes ($m_1$ and $m_2$) in terms of $a$.

2. Find the coordinates of the intersection points A and B of the line $y=2x+4$ with the lines $y=m_1x$ and $y=m_2x$.

3. Calculate the area of triangle AOB using the coordinates of O, A, and B. The area formula involves $m_1$, $m_2$, and $m=2$.

4. The problem statement is incomplete, likely missing a condition. Assuming the line $y=2x+4$ is an angle bisector of the pair of lines represented by the curve.

5. Find the equation of the angle bisectors of the pair of lines in terms of $a$.

6. The slope of the angle bisector must be 2. Use this condition to find the value of $a$.

7. Substitute the value of $a$ back into the expressions for $m_1+m_2$ and the area to find their values.

8. Verify the result by finding $m_1$ and $m_2$ for the determined value of $a$, finding the coordinates of A and B, and calculating the area directly.

The final answer is $\boxed{m_1+m_2 = \frac{20}{3}, Area = \frac{16}{5}}$.