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(B) The equation of the parabola is $y^2=4ax$. The focus is $(a, 0)$ and the directrix is $x=-a$.The distance between the focus and the directrix is $

Vedclass · Accepted Answer

$2x-y=9$

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(B) The equation of the parabola is $y^2=4ax$. The focus is $(a, 0)$ and the directrix is $x=-a$.The distance between the focus and the directrix is $2a$.Given $2a = \frac{3}{2}$,so $a = \frac{3}{4}$.The point on the parabola is $(4a, -4a) = (4 	imes \frac{3}{4}, -4 	imes \frac{3}{4}) = (3, -3)$.The equation of the normal to the parabola $y^2=4ax$ at $(x_1, y_1)$ is $y-y_1 = -\frac{y_1}{2a}(x-x_1)$.Substituting $x_1=3, y_1=-3$ and $a=\frac{3}{4}$:$y - (-3) = -\frac{-3}{2(3/4)}(x-3)$$y+3 = \frac{3}{3/2}(x-3)$$y+3 = 2(x-3)$$y+3 = 2x-6$$2x-y=9$.

If the perpendicular distance from the focus of a parabola $y^2=4ax$ to its directrix is $\frac{3}{2}$,then the equation of the normal drawn at $(4a, -4a)$ is

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