A point on the parabola whose focus is S(1,-1 | vedclass

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(A) The axis of the parabola is the line passing through the focus $S(1,-1)$ and vertex $A(1,1)$. Since the $x$-coordinates are the same,the axis is t

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$\left(3, \frac{1}{2}\right)$

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(A) The axis of the parabola is the line passing through the focus $S(1,-1)$ and vertex $A(1,1)$. Since the $x$-coordinates are the same,the axis is the vertical line $x=1$.The distance between the vertex $A(1,1)$ and focus $S(1,-1)$ is $a = \sqrt{(1-1)^2 + (1 - (-1))^2} = 2$.Since the vertex is above the focus,the parabola opens downwards. The directrix is a horizontal line at a distance $a=2$ above the vertex $A(1,1)$.The equation of the directrix is $y = 1 + 2 = 3$.By the definition of a parabola,for any point $P(x,y)$ on it,the distance to the focus equals the distance to the directrix: $PS^2 = PM^2$.$(x-1)^2 + (y+1)^2 = (y-3)^2$.$(x-1)^2 + y^2 + 2y + 1 = y^2 - 6y + 9$.$(x-1)^2 = -8y + 8 = 8(1-y)$.Checking option $A$: For $\left(3, \frac{1}{2}\right)$,$(3-1)^2 = 2^2 = 4$ and $8(1 - \frac{1}{2}) = 8(\frac{1}{2}) = 4$.Since $4=4$,the point $\left(3, \frac{1}{2}\right)$ lies on the parabola.

$A$ point on the parabola whose focus is $S(1,-1)$ and whose vertex is $A(1,1)$ is

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