Question 10,11
he parabola
Part 1 of 3
Find the equation of the parabola described below. Find the two points that define the latus rectum, and graph the equatio Focus at $(-3,-5)$, directrix the line $x=1$
The equation of the parabola is
(Type an equation. Use integers or fractions for any numbers in the equation)

Step 1 ：The equation of a parabola with a focus at \((h,k)\) and a directrix at \(x = d\) is given by \((y - k)^2 = 4a(x - h)\), where \(a\) is the distance from the focus to the directrix.

Step 2 ：In this case, the focus is at \((-3,-5)\) and the directrix is at \(x = 1\), so \(a = -3 - 1 = -4\).

Step 3 ：Therefore, the equation of the parabola is \((y + 5)^2 = -16(x + 3)\).

Step 4 ：The latus rectum of a parabola is a line segment that passes through the focus and is perpendicular to the axis of symmetry. The length of the latus rectum is \(4a\), so in this case it is \(4 * -4 = -16\).

Step 5 ：The two points that define the latus rectum can be found by moving \(-16/2 = -8\) units up and down from the focus along the axis of symmetry. The axis of symmetry is the line \(y = -5\), so the two points are \((-3, -5 - 8) = (-3, -13)\) and \((-3, -5 + 8) = (-3, 3)\).

Step 6 ：Final Answer: The equation of the parabola is \((y + 5)^2 = -16(x + 3)\). The two points that define the latus rectum are \((-3, -13)\) and \((-3, 3)\).