Let P be a parabola with vertex (2,3) and dir | vedclass

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(D) The axis of the parabola is perpendicular to the directrix $2x+y=6$ and passes through the vertex $(2,3)$.Slope of directrix is $-2$,so slope of a

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$\frac{656}{25}$

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(D) The axis of the parabola is perpendicular to the directrix $2x+y=6$ and passes through the vertex $(2,3)$.Slope of directrix is $-2$,so slope of axis is $\frac{1}{2}$.Equation of axis: $y-3 = \frac{1}{2}(x-2) \Rightarrow x-2y+4=0$.The intersection of the axis and directrix is the point $Z$. Solving $2x+y=6$ and $x-2y=-4$,we get $Z = (1.6, 2.8)$.Let the focus be $S(\alpha, \beta)$. Since the vertex $V(2,3)$ is the midpoint of $SZ$,we have $\frac{\alpha+1.6}{2} = 2 \Rightarrow \alpha = 2.4$ and $\frac{\beta+2.8}{2} = 3 \Rightarrow \beta = 3.2$.So,the focus is $(2.4, 3.2)$.The ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through $(2.4, 3.2)$,so $\frac{(2.4)^2}{a^2} + \frac{(3.2)^2}{b^2} = 1$.Given $e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2}$ $\Rightarrow b^2 = \frac{a^2}{2}$ $\Rightarrow a^2 = 2b^2$.Substituting $a^2=2b^2$ into the ellipse equation: $\frac{5.76}{2b^2} + \frac{10.24}{b^2} = 1$ $\Rightarrow \frac{2.88+10.24}{b^2} = 1$ $\Rightarrow b^2 = 13.12 = \frac{1312}{100} = \frac{328}{25}$.Then $a^2 = 2 	imes \frac{328}{25} = \frac{656}{25}$.The length of the latus rectum is $L = \frac{2b^2}{a}$. The square of the length is $L^2 = \frac{4b^4}{a^2} = \frac{4b^4}{2b^2} = 2b^2 = 2 	imes \frac{328}{25} = \frac{656}{25}$.

Let $P$ be a parabola with vertex $(2,3)$ and directrix $2x+y=6$. Let an ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ of eccentricity $e=\frac{1}{\sqrt{2}}$ pass through the focus of the parabola $P$. Then the square of the length of the latus rectum of $E$ is:

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