Let the abscissae of the two points \(P\) and \(Q\) be the roots of \(2x^2 – rx + p = 0\) and the ordinates of \(P\) and \(Q\) be the roots of \(x^2 – sx – q = 0\). If the equation of the circle described on \(PQ\) as diameter is \(2(x^2 + y^2) – 11x – 14y – 22 = 0\), then \(2r + s – 2q + p\) is equal to _________.
Correct Answer: 7
Solution and Explanation
Suppose P(x1, y1) & Q(x2, y2)
Solve: 2x2 – rx + p = 0 for x1 and x2
Solve: x2 – sx – q = 0 for y1 and y2
So, the quation of circle = (x – x1) (x – x2) + (y – y1) (y – y2) = 0
x2 – (x1 + x2)x + x1x2 + y2 – (y1 + y2)y + y1y2 = 0
x2−\(\frac r2\)x+\(\frac p2\)+y2+sy−q = 0
2x2 + 2y2 – rx + 2sy + p – 2q = 0
On comparing with 2x2 + 2y2 – 11x – 14y – 22 = 0
r = 11,
s = 7,
p – 2q = –22
Then,
2r + s + p – 2q
= 22 + 7 – 22
= 7
So, the answer is 7.
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