The equation ax² - 6xy + y² + 2gx + 2fy + c = 0 represents a pair of straight lines whose slopes are m and m². Find the sum of all possible values of a.

Correct answer: -19

Solution

From the homogeneous part, slopes satisfy m² - 6m + a = 0 (dividing y² coefficient = 1). By Vieta's: sum of slopes = 6 and product = a. Solving m² + m - 6 = 0 gives m = 2 or m = -3, yielding a = 8 or a = -27, with sum -19.

Related JEE Advanced Maths questions

Determine the range of β for which the point (0, β) is located on or within the triangle formed by the equations y + 3x + 2 = 0, 3y − 2x − 5 = 0, and 4y − x − 14 = 0.
For the equation a² + b² − c² − 2ab = 0, the common intersection point of the set of lines represented by ax + by + c = 0 is located on which line?
If the points (a³/(a − 3), (a² − 3)/(a − 1)), (b³/(b − 3), (b² − 3)/(b − 1)), and (c³/(c − 3), (c² − 3)/(c − 1)), where a, b, and c are not equal to 1, are situated on the line l₁x + m₁y + n₁ = 0, then which of the following is true?
In triangle ABC, the median AD is divided into two equal parts at point E. Line BE intersects AC at point F. What is the ratio of AF to AC?
Let m1 and m2 be the slopes of two adjacent sides of a square of side length a such that a² + 11a + 3(m1² + m2²) = 220. One vertex of the square is at the point (10(cos(alpha) - sin(alpha)), 10(sin(alpha) + cos(alpha))), where alpha is in (0, pi/2), and one diagonal of the square has equation (cos(alpha) - sin(alpha))x + (sin(alpha) + cos(alpha))y = 10. Find the value of 72(sin⁴(alpha) + cos⁴(alpha)) + a² - 3a + 13.
Let P be any point on the line x - y + 3 = 0 and let A = (3, 4) be a fixed point. A family of lines given by (3 sec(t) + 5 cosec(t)) x + (7 sec(t) - 3 cosec(t)) y + 11(sec(t) - cosec(t)) = 0 passes through a fixed point B for all permissible values of t. If the maximum value of |PA - PB| is expressed as 2 * sqrt(2n) where n is a natural number, find n.

⚔️ Practice JEE Advanced Maths free + battle 1v1 →