If the focus of a parabola is (3, -4) and its | vedclass

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(D) The axis of the parabola is perpendicular to the directrix $x + y - 2 = 0$. Thus,the equation of the axis is $x - y + k = 0$. Since it passes thro

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$(15/4, -13/4)$

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(D) The axis of the parabola is perpendicular to the directrix $x + y - 2 = 0$. Thus,the equation of the axis is $x - y + k = 0$. Since it passes through the focus $S(3, -4)$,we have $3 - (-4) + k = 0$,which gives $k = -7$. The equation of the axis is $x - y - 7 = 0$. Let $Z$ be the intersection of the axis and the directrix. Solving $x + y - 2 = 0$ and $x - y - 7 = 0$,we get $Z = (9/2, -5/2)$. The vertex $A$ is the midpoint of $SZ$. $A = \left( \frac{3 + 9/2}{2}, \frac{-4 - 5/2}{2} \right) = \left( \frac{15/2}{2}, \frac{-13/2}{2} \right) = \left( \frac{15}{4}, -\frac{13}{4} \right)$.

If the focus of a parabola is $(3, -4)$ and its directrix is $x + y - 2 = 0$,find the vertex.

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