If the focal chord of the parabola x^2=12y dr | vedclass

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(C) The parabola is $x^2=12y$,which is of the form $x^2=4ay$,where $4a=12$,so $a=3$. The focus is $(0,a) = (0,3)$.Since the chord passes through the f

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$\frac{1}{3}$

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(C) The parabola is $x^2=12y$,which is of the form $x^2=4ay$,where $4a=12$,so $a=3$. The focus is $(0,a) = (0,3)$.Since the chord passes through the focus $(0,3)$ and the point $(3,0)$,its equation is $\frac{x}{3} + \frac{y}{3} = 1$,which simplifies to $y = 3-x$.Substituting $y = 3-x$ into the parabola equation $x^2 = 12y$:$x^2 = 12(3-x)$$x^2 = 36 - 12x$$x^2 + 12x - 36 = 0$.Let the abscissae of points $P$ and $Q$ be $x_1$ and $x_2$. These are the roots of the quadratic equation $x^2 + 12x - 36 = 0$.From the properties of roots,the sum of roots $x_1 + x_2 = -12$ and the product of roots $x_1 x_2 = -36$.The sum of the reciprocals of the abscissae is $\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2}$.Substituting the values: $\frac{-12}{-36} = \frac{1}{3}$.

If the focal chord of the parabola $x^2=12y$ drawn through the point $(3,0)$ intersects the parabola at the points $P$ and $Q$,then the sum of the reciprocals of the abscissae of the points $P$ and $Q$ is

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