Consider the parabola with vertex left(frac{1 | vedclass

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(B) The equation of the parabola with vertex $(h, k) = \left(\frac{1}{2}, \frac{3}{4}\right)$ and directrix $y = k - a = \frac{1}{2}$ is given by $(x

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$\frac{125}{16}$

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(B) The equation of the parabola with vertex $(h, k) = \left(\frac{1}{2}, \frac{3}{4}ight)$ and directrix $y = k - a = \frac{1}{2}$ is given by $(x - h)^2 = 4a(y - k)$.Since $k - a = \frac{1}{2}$,we have $\frac{3}{4} - a = \frac{1}{2}$,so $a = \frac{1}{4}$.The equation is $\left(x - \frac{1}{2}ight)^2 = 4 	imes \frac{1}{4} \left(y - \frac{3}{4}ight)$,which simplifies to $\left(x - \frac{1}{2}ight)^2 = y - \frac{3}{4}$.For $x = -\frac{1}{2}$,we have $\left(-\frac{1}{2} - \frac{1}{2}ight)^2 = y - \frac{3}{4}$ $\Rightarrow 1 = y - \frac{3}{4}$ $\Rightarrow y = \frac{7}{4}$. Thus,$P = \left(-\frac{1}{2}, \frac{7}{4}ight)$.Differentiating the parabola equation: $2\left(x - \frac{1}{2}ight) = \frac{dy}{dx}$.At $x = -\frac{1}{2}$,the slope of the tangent $m_T = 2\left(-\frac{1}{2} - \frac{1}{2}ight) = -2$.The slope of the normal $m_N = -\frac{1}{m_T} = \frac{1}{2}$.The equation of the normal at $P$ is $y - \frac{7}{4} = \frac{1}{2} \left(x + \frac{1}{2}ight) \Rightarrow y = \frac{x}{2} + 2$.Substituting $y = \frac{x}{2} + 2$ into the parabola equation: $\left(x - \frac{1}{2}ight)^2 = \left(\frac{x}{2} + 2ight) - \frac{3}{4}$ $\Rightarrow x^2 - x + \frac{1}{4} = \frac{x}{2} + \frac{5}{4}$.$x^2 - \frac{3}{2}x - 1 = 0$ $\Rightarrow 2x^2 - 3x - 2 = 0$ $\Rightarrow (2x + 1)(x - 2) = 0$.Since $x = -\frac{1}{2}$ corresponds to $P$,the point $Q$ has $x = 2$. Then $y = \frac{2}{2} + 2 = 3$,so $Q = (2, 3)$.$(PQ)^2 = \left(2 - (-\frac{1}{2})ight)^2 + \left(3 - \frac{7}{4}ight)^2 = \left(\frac{5}{2}ight)^2 + \left(\frac{5}{4}ight)^2 = \frac{25}{4} + \frac{25}{16} = \frac{100 + 25}{16} = \frac{125}{16}$.

Consider the parabola with vertex $\left(\frac{1}{2}, \frac{3}{4}\right)$ and the directrix $y=\frac{1}{2}$. Let $P$ be the point where the parabola meets the line $x=-\frac{1}{2}$. If the normal to the parabola at $P$ intersects the parabola again at the point $Q$,then $(PQ)^{2}$ is equal to :

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