Find the equation in standard form of the parabola that has vertex $(1,-5)$, has a horizontal axis of symmetry, and passes through the point $(-3,-3)$.
[[fieldinquestionpre]]
a. $(y+5)^{2}=-16(x-1)$
b. $(y+1)^{2}=16(x-5)$
c. $(y-1)^{2}=16(x+5)$
d. $(y+1)^{2}=-16(x-5)$
e. $(y+5)^{2}=16(x-1)$

Step 1 ：The standard form of a parabola with a horizontal axis of symmetry is \((y-k)^2=4p(x-h)\), where \((h,k)\) is the vertex of the parabola and \(p\) is the distance from the vertex to the focus or the directrix. Since the vertex is given as \((1,-5)\), the equation of the parabola can be written as \((y+5)^2=4p(x-1)\).

Step 2 ：We also know that the parabola passes through the point \((-3,-3)\). Substituting this point into the equation will allow us to solve for \(p\).

Step 3 ：Solving the equation \((y + 5)^2 = 4p(x - 1)\) for \(p\) when \(x = -3\) and \(y = -3\) gives \(p = -1/4\).

Step 4 ：The value of \(p\) is \(-1/4\). This means the parabola opens to the left since \(p\) is negative and the axis of symmetry is horizontal.

Step 5 ：We can substitute \(p\) back into the equation of the parabola to get the final equation \((y+5)^{2}=-4(x-1)\).

Step 6 ：Multiplying both sides of the equation by 4 gives the final equation in standard form: \((y+5)^{2}=-16(x-1)\).

Step 7 ：Final Answer: The equation of the parabola in standard form is \(\boxed{(y+5)^{2}=-16(x-1)}\).