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(A) The equation of the parabola is $y^{2} = 12x$,which is of the form $y^{2} = 4ax$,where $a = 3$.For any point $(x_{1}, y_{1})$,the pair of tangents

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$90$

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(A) The equation of the parabola is $y^{2} = 12x$,which is of the form $y^{2} = 4ax$,where $a = 3$.For any point $(x_{1}, y_{1})$,the pair of tangents is given by $SS_{1} = T^{2}$.Here,$S = y^{2} - 12x$,$S_{1} = (2)^{2} - 12(-3) = 4 + 36 = 40$,and $T = y(2) - 6(x - 3) = 2y - 6x + 18$.Substituting these into the formula:$(y^{2} - 12x)(40) = (2y - 6x + 18)^{2}$$40y^{2} - 480x = 4(y - 3x + 9)^{2}$$10y^{2} - 120x = y^{2} + 9x^{2} + 81 - 6xy - 54x + 18y$$9x^{2} - 9y^{2} - 6xy + 66x + 18y + 81 = 0$.For a general second-degree equation $Ax^{2} + 2Hxy + By^{2} + 2Gx + 2Fy + C = 0$,the angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{H^{2} - AB}}{A + B} ight|$.Here,$A = 9$,$B = -9$,and $H = -3$.Since $A + B = 9 + (-9) = 0$,the lines are perpendicular.Therefore,the angle between the tangents is $90^{\circ}$.

The angle between the tangents drawn to the parabola $y^{2}=12x$ from the point $(-3, 2)$ is (in $^{\circ}$)

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