Let the abscissae of the two points P and Q be the roots of 2 x 2 - r x + p = 0 and the ordinates of P and Q be the roots of x 2 - s x - q = 0 . If the equation of the circle described on P Q as diameter is 2 x 2 + y 2 - 11 x - 14 y - 22 = 0 , then 2 r + s - 2 q + p is equal to ______.

Step-by-step solution

Let the roots of 2 x 2 - r x + p = 0 are x 1 ,   x 2 and roots of y 2 - s y - q = 0 are y 1 ,   y 2 So, x 1 + x 2 = r 2 ,   x 1 x 2 = p 2 ,   y 1 + y 2 = s ,   y 1 y 2 = - q Equation of the circle with P Q as diameter will be x - x 1 x - x 2 + y - y 1 y - y 2 = 0 i.e. 2 x 2 + y 2 - r x - 2 s y + p - 2 q = 0 On comparing with the given equation r = 11 , s = 7 p - 2 q = - 22 ∴   2 r + s - 2 q + p = 22 + 7 - 22 = 7