Question

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Explanation:

Step1: Define parabola distance rule

A point $(x,y)$ on the parabola satisfies the equality of the distance to the focus $(3,1)$ and the distance to the directrix $y=5$. The distance to the focus is $\sqrt{(x-3)^2+(y-1)^2}$, and the distance to the directrix is $|y-5|$. Set them equal:
$$\sqrt{(x-3)^2+(y-1)^2} = |y-5|$$

Step2: Simplify to standard parabola equation

Square both sides to eliminate the square root and absolute value:
$$(x-3)^2+(y-1)^2=(y-5)^2$$
Expand both sides:
$$(x-3)^2 + y^2 -2y +1 = y^2 -10y +25$$
Cancel $y^2$, then rearrange terms:
$$(x-3)^2 = -8y +24$$
Simplify to:
$$(x-3)^2 = -8(y-3)$$

Step3: Test each point

Test A. $(-5,-3)$

Substitute $x=-5, y=-3$ into the left side:
$$(-5-3)^2=(-8)^2=64$$
Substitute into the right side:
$$-8(-3-3)=-8(-6)=48$$
$64
eq 48$, so the point is not on the parabola.

Test B. $(-1,0)$

Substitute $x=-1, y=0$ into the left side:
$$(-1-3)^2=(-4)^2=16$$
Substitute into the right side:
$$-8(0-3)=-8(-3)=24$$
$16
eq 24$, so the point is not on the parabola.

Test C. $(3,1)$

Substitute $x=3, y=1$ into the left side:
$$(3-3)^2=0^2=0$$
Substitute into the right side:
$$-8(1-3)=-8(-2)=16$$
$0
eq 16$, so the point is not on the parabola.

Test D. $(3,3)$

Substitute $x=3, y=3$ into the left side:
$$(3-3)^2=0^2=0$$
Substitute into the right side:
$$-8(3-3)=-8(0)=0$$
$0=0$, so the point is on the parabola.

Test E. $(7,1)$

Substitute $x=7, y=1$ into the left side:
$$(7-3)^2=(4)^2=16$$
Substitute into the right side:
$$-8(1-3)=-8(-2)=16$$
$16=16$, so the point is on the parabola.

Answer:

D. (3,3), E. (7,1)