The line y = mx + 1 cuts the parabola $y^{2} = 4 a x$ at two points P and Q. If equation of circle with diameter as PQ is $x^{2} + y^{2} + 18 x + 4 y + 5 = 0$ then a + m equal to

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Detailed Solution

$\begin{matrix} b y e l e m i n a t i n g y f r o m t h e g i v i e n e q u a t i o n s {(m x + 1)}^{2} = 4 a x \\ m^{2} x^{2} + 2 m x + 1 = 4 a x \\ m^{2} x^{2} + (2 m - 4 a) x + 1 = 0 \\ (x - x_{1}) (x - x_{2}) = x^{2} + \frac{2 x (m - 2 a)}{m^{2}} + \frac{1}{m^{2}} ...... (1) \\ b y e l e m i n a t i n g x f r o m t h e g i v e n e q u a t i o n s y^{2} - 4 a (\frac{y - 1}{m}) = 0 \\ (y - y_{1}) (y - y_{2}) = y^{2} - \frac{4 a y}{m} + \frac{4 a}{m} ...... (2) \\ Add the above two equations \\ (x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = x^{2} + y^{2} + \frac{2 x (m - 2 a)}{m^{2}} - \frac{4 a y}{m} + \frac{4 a}{m} + \frac{1}{m^{2}} \\ = x^{2} + y^{2} + 18 x + 4 y + 5 \\ m = \frac{1}{3}, a = - \frac{1}{3} \end{matrix}$