Let Z be the point of intersection of the axi | vedclass

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(A) The given equation of the parabola is $4x^2 - 12x + 4y + 5 = 0$.Dividing by $4$,we get $x^2 - 3x + y + \frac{5}{4} = 0$.Completing the square for

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$\left(\frac{3}{2}, \frac{13}{12}\right)$

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(A) The given equation of the parabola is $4x^2 - 12x + 4y + 5 = 0$.Dividing by $4$,we get $x^2 - 3x + y + \frac{5}{4} = 0$.Completing the square for $x$: $(x - \frac{3}{2})^2 = -y - \frac{5}{4} + \frac{9}{4} = -(y - 1)$.This is of the form $(x - h)^2 = -4a(y - k)$,where $h = \frac{3}{2}$,$k = 1$,and $4a = 1 \Rightarrow a = \frac{1}{4}$.The vertex is $V = (h, k) = (\frac{3}{2}, 1)$.The focus $S$ is $(h, k - a) = (\frac{3}{2}, 1 - \frac{1}{4}) = (\frac{3}{2}, \frac{3}{4})$.The directrix is $y = k + a = 1 + \frac{1}{4} = \frac{5}{4}$.The axis of the parabola is $x = h = \frac{3}{2}$.The point $Z$ is the intersection of the axis and the directrix,so $Z = (\frac{3}{2}, \frac{5}{4})$.We need to find the point $P$ that divides $SZ$ in the ratio $2:1$. Using the section formula:$P = \left(\frac{2 	imes \frac{3}{2} + 1 	imes \frac{3}{2}}{2+1}, \frac{2 	imes \frac{5}{4} + 1 	imes \frac{3}{4}}{2+1}ight) = \left(\frac{3 + \frac{3}{2}}{3}, \frac{\frac{5}{2} + \frac{3}{4}}{3}ight) = \left(\frac{9/2}{3}, \frac{13/4}{3}ight) = \left(\frac{3}{2}, \frac{13}{12}ight)$.

Let $Z$ be the point of intersection of the axis and the directrix of the parabola $4x^2 - 12x + 4y + 5 = 0$. If $S$ is its focus,then the point which divides $SZ$ in the ratio $2:1$ is

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