Let one root of the quadratic equation in x : (k^2 - 15k + 27)x^2 + 9(k-1)x + 18 = 0 be twice the other. Then the length of the latus rectum of the parabola y^2 = 6kx is equal to:

Step-by-step solution

Let the roots of the given quadratic equation be and 2 . Sum of the roots: + 2 = -9(k-1) k^2 - 15k + 27 3 = -9(k-1) k^2 - 15k + 27 = -3(k-1) k^2 - 15k + 27 Product of the roots: 2 = 18 k^2 - 15k + 27 2 ^2 = 18 k^2 - 15k + 27 ^2 = 9 k^2 - 15k + 27 Substituting the value of from the sum into the product equation: ( -3(k-1) k^2 - 15k + 27 )^2 = 9 k^2 - 15k + 27 9(k-1)^2 (k^2 - 15k + 27)^2 = 9 k^2 - 15k + 27 Since k^2 - 15k + 27 0 , we can simplify to: (k-1)^2 = k^2 - 15k + 27 k^2 - 2k + 1 = k^2 - 15k + 27 13k = 26 k =