A chord is drawn through the focus of the par | vedclass

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Question

(A) The equation of the parabola is $y^2 = 6x$,which is of the form $y^2 = 4ax$ with $4a = 6$,so $a = \frac{3}{2}$.The focus is $S = (\frac{3}{2}, 0)$

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$\frac{\sqrt{5}}{2}$

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(A) The equation of the parabola is $y^2 = 6x$,which is of the form $y^2 = 4ax$ with $4a = 6$,so $a = \frac{3}{2}$.The focus is $S = (\frac{3}{2}, 0)$ and the vertex is $V = (0, 0)$.Let the equation of the chord passing through the focus be $y - 0 = m(x - \frac{3}{2})$,which simplifies to $mx - y - \frac{3m}{2} = 0$.The distance $d$ of this line from the vertex $(0, 0)$ is given by $d = \frac{|m(0) - 0 - \frac{3m}{2}|}{\sqrt{m^2 + (-1)^2}} = \frac{|-\frac{3m}{2}|}{\sqrt{m^2 + 1}}$.Given $d = \frac{\sqrt{5}}{2}$,we have $\frac{|\frac{3m}{2}|}{\sqrt{m^2 + 1}} = \frac{\sqrt{5}}{2}$.Squaring both sides,we get $\frac{9m^2}{4(m^2 + 1)} = \frac{5}{4}$.$9m^2 = 5(m^2 + 1)$ $\Rightarrow 9m^2 = 5m^2 + 5$ $\Rightarrow 4m^2 = 5$.$m^2 = \frac{5}{4} \Rightarrow m = \pm \frac{\sqrt{5}}{2}$.Thus,the slope can be $\frac{\sqrt{5}}{2}$.

$A$ chord is drawn through the focus of the parabola $y^2 = 6x$ such that its distance from the vertex of this parabola is $\frac{\sqrt{5}}{2}$. Then,its slope can be:

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