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(B) The equation of the parabola is $y^2 = 4ax$,where $4a = 7$,so $a = \frac{7}{4}$.The equation of a normal to the parabola $y^2 = 4ax$ at point $(at

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(B) The equation of the parabola is $y^2 = 4ax$,where $4a = 7$,so $a = \frac{7}{4}$.The equation of a normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.Since the normal passes through the point $(2, 0)$,we substitute $x = 2$ and $y = 0$ into the equation:$0 = -t(2) + 2(\frac{7}{4})t + \frac{7}{4}t^3$$0 = -2t + \frac{7}{2}t + \frac{7}{4}t^3$$0 = \frac{3}{2}t + \frac{7}{4}t^3$$0 = t(\frac{3}{2} + \frac{7}{4}t^2)$This gives $t = 0$ or $t^2 = -\frac{3}{2} 	imes \frac{4}{7} = -\frac{6}{7}$.Since $t^2$ cannot be negative for real $t$,the only real solution is $t = 0$.Thus,only $1$ normal can be drawn.

The number of normals that can be drawn through the point $(2,0)$ to the parabola $y^2=7x$ is

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