-A- circle is described on a focal chord as diameter ; if m be the tangent of the inclination of the chord to the axis, prove that the equation to the circle is
$x^{2} + y^{2} - 2 a x (1 + \frac{2}{m^{2}}) - \frac{4 a y}{m} - 3 a^{2} = 0 .$

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Solution

Let the end points of the focal chord be $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$

Slope of $P Q = \frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}}$

$\frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}} = m \frac{2 a (t_{2} - t_{1})}{a (t_{2} - t_{1}) (t_{2} + t_{1})} = m \Rightarrow t_{1} + t_{2} = \frac{2}{m} . . . . . . (i)$

Equation of circle is

$(x - a t_{1}^{2}) (x - a t_{2}^{2}) + (y - 2 a t_{1}) (y - 2 a t_{2}) = 0 x^{2} - a x (t_{1}^{2} + t_{2}^{2}) + a^{2} t_{1}^{2} t_{2}^{2} + y^{2} - 2 a y (t_{1} + t_{2}) + 4 a^{2} t_{1} t_{2} = 0$

$x^{2} - a x ((t_{1} + t_{2})^{2} - 2 t_{1} t_{2}) + a^{2} t_{1}^{2} t_{2}^{2} + y^{2} - 2 a y (t_{1} + t_{2}) + 4 a^{2} t_{1} t_{2} = 0$

For focal chord $t_{1} t_{2} = - 1$

$x^{2} - a x ((t_{1} + t_{2})^{2} + 2) + y^{2} - 2 a y (t_{1} + t_{2}) - 3 a^{2} = 0$

Substituting $(i)$

$x^{2} - a x (2 + \frac{4}{m^{2}}) + y^{2} - 2 a y \frac{2}{m} - 3 a^{2} = 0 x^{2} - 2 a x (1 + \frac{2}{m^{2}}) + y^{2} - \frac{4 a y}{m} - 3 a^{2} = 0$

Hence proved.

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