Answer:
(a)
• The vertex of the parabola, (h,k)=(0,0)
,
• The value of p = -3
• The focus is at (0,-3).
,
• The focal diameter is 12
(b)The endpoints of latus rectum are (-1/12, -1/6) and (-1/12, 1/6).
(c)See Graph below
(d)
• I. The equation for the directrix is y=3.
,
• II. The axis of symmetry is at x=0.
Explanation:
Given the equation of the parabola:
[tex]x^2=-12y[/tex]
For an up-facing parabola with vertex at (h, k) and a focal length Ipl, the standard equation is:
[tex](x-h)^2=4p(y-k)[/tex]
Rewrite the equation in the given format:
[tex]\begin{gathered} (x-0)^2=4(-3)(y-0) \\ \implies(h,k)=(0,0) \\ \implies p=-3 \end{gathered}[/tex]
• The vertex of the parabola, (h,k)=(0,0)
,
• The value of p = -3
The focus is calculated using the formula:
[tex]\begin{gathered} (h,k+p) \\ \implies Focus=(0,0-3)=(0-3) \end{gathered}[/tex]
• The focus is at (0,-3).
Focal Diameter
Comparing the given equation with x²=4py, we have:
[tex]\begin{gathered} x^2=4ay \\ x^2=-12y \\ 4a=-12 \\ \implies a=-3 \\ \text{ Focal Diameter =4\mid a\mid=4\mid3\mid=12} \end{gathered}[/tex]
The focal diameter is 12
Part B (The endpoints of the latus rectum).
First, rewrite the equation in the standard form:
[tex]\begin{gathered} y=-\frac{1}{12}x^2 \\ \implies a=-\frac{1}{12} \end{gathered}[/tex]
The endpoints are:
[tex]\begin{gathered} (a,2a)=(-\frac{1}{12},-\frac{1}{6}) \\ (a,-2a)=(-\frac{1}{12},\frac{1}{6}) \end{gathered}[/tex]
The endpoints of latus rectum are (-1/12, -1/6) and (-1/12, 1/6).
Part C
The graph of the parabola is given below:
Part D
I. The equation for the directrix is of the form y=k-p.
[tex]\begin{gathered} y=0-(-3) \\ y=3 \end{gathered}[/tex]
The equation for the directrix is y=3.
II. The axis of symmetry is the x-value at the vertex.
The axis of symmetry is at x=0.
