Question

for the equation of the parabola given in the form $(y - k)^2 = 4p(x - h)$,
(a) identify the vertex, value of $p$, focus, and focal diameter of the parabola.
(b) identify the endpoints of the latus rectum.
(c) graph the parabola.
(d) write equations for the directrix and axis of symmetry.
express numbers in exact, simplest form.
$(y + 4)^2 = -16(x + 2)$

Explanation:

Step1: Identify vertex, \(p\), focus, focal diameter

The standard form of a parabola is \((y - k)^2=4p(x - h)\).
For the equation \((y + 4)^2=-16(x + 2)\), we have \(h=-2\), \(k = - 4\).
Comparing with \((y - k)^2=4p(x - h)\), we get \(4p=-16\), so \(p=-4\).
The vertex \((h,k)=(-2,-4)\).
The focus \((h + p,k)=(-2-4,-4)=(-6,-4)\).
The focal diameter \(|4p| = 16\).

Step2: Find endpoints of latus rectum

The latus - rectum is parallel to the \(y\) - axis (since the parabola opens to the left).
The endpoints of the latus rectum:
Let \(x=h + p=-6\). Substitute into \((y + 4)^2=-16(x + 2)\), when \(x=-6\), \((y + 4)^2=-16(-6 + 2)=64\), \(y+4=\pm8\), \(y = 4\) or \(y=-12\).
The endpoints are \((-6,4)\) and \((-6,-12)\).

Step3: Find directrix and axis of symmetry

The directrix is \(x=h - p\). Substituting \(h=-2\) and \(p=-4\), we get \(x=-2-(-4)=2\).
The axis of symmetry is \(y = k\), so \(y=-4\).

Step4: Graph the parabola

• Plot the vertex \((-2,-4)\).
• Plot the focus \((-6,-4)\).
• Draw the directrix \(x = 2\).
• Plot the endpoints of the latus rectum \((-6,4)\) and \((-6,-12)\).
• Sketch the parabola opening to the left.

Answer:

(a) Vertex: \((-2,-4)\), \(p=-4\), Focus: \((-6,-4)\), Focal diameter: \(16\)
(b) Endpoints of latus rectum: \((-6,4)\) and \((-6,-12)\)
(c) Graph as described above
(d) Directrix: \(x = 2\), Axis of symmetry: \(y=-4\)