Consider an ellipse E,a hyperbola H,and a par | vedclass

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(A) For any conic section with focus $S$ and directrix $L$,the vertex $V$ lies on the axis of the conic section,which is the line passing through the

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$\alpha > \beta > \gamma$

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(A) For any conic section with focus $S$ and directrix $L$,the vertex $V$ lies on the axis of the conic section,which is the line passing through the focus $S(2, 3)$ and perpendicular to the directrix $x + y - 10 = 0$.The slope of the directrix is $-1$,so the slope of the axis is $1$. The equation of the axis is $y - 3 = 1(x - 2)$,which simplifies to $y = x + 1$.The intersection of the axis $y = x + 1$ and the directrix $x + y = 10$ is found by $x + (x + 1) = 10$,giving $2x = 9$,so $x = 4.5$ and $y = 5.5$. Let this point be $Z(4.5, 5.5)$.The distance $SZ = \sqrt{(4.5 - 2)^2 + (5.5 - 3)^2} = \sqrt{2.5^2 + 2.5^2} = 2.5\sqrt{2}$.For a conic with eccentricity $e$,the distance from the focus to the vertex is $SV = \frac{e}{1+e} SZ$ (for the vertex between focus and directrix).For an ellipse,$0 For a parabola,$e = 1$,so $SV_P = \frac{1}{2} SZ$.For a hyperbola,$e > 1$,so $SV_H = \frac{e_H}{1+e_H} SZ$. As $e_H$ increases,$SV_H$ approaches $SZ$.Since the vertex is between the focus and the directrix,the distance from the directrix to the vertex is $VZ = SZ - SV = SZ - \frac{e}{1+e} SZ = \frac{1}{1+e} SZ$.For the ellipse $(e  \frac{1}{2} SZ$.For the parabola $(e = 1)$,$VZ_P = \frac{1}{2} SZ$.For the hyperbola $(e > 1)$,$VZ_H = \frac{1}{1+e_H} SZ Thus,$VZ_E > VZ_P > VZ_H$. Since the vertices lie on the same line,the coordinates follow the order $\alpha > \beta > \gamma$.

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