Question

1. write the equation for the parabola with vertex (0, 6) and focus (0, 3)

for numbers 10 and 11 name:
a) the coordinates of the vertex and focus
b) the equation of the axis of symmetry and directrix
c) the direction of opening
d) length of the latus rectum

1. $(y - 6)^2 = -8 (x - 6)$
2. $y = \frac{1}{2} x^2 + 12x - 8$

Explanation:

Problem 9:

Step 1: Identify the form of the parabola

The vertex is \((0, 6)\) and the focus is \((0, 3)\). Since the \(x\)-coordinates of the vertex and focus are the same (\(x = 0\)), the parabola opens up or down. The standard form of a parabola that opens up or down is \((x - h)^2 = 4p(y - k)\), where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus (and also from the vertex to the directrix).

Step 2: Find the value of \(p\)

The vertex is \((h, k)=(0, 6)\) and the focus is \((0, 3)\). The distance between the vertex \((0, 6)\) and the focus \((0, 3)\) is \(|6 - 3| = 3\). Since the focus is below the vertex, \(p\) is negative. So \(p=- 3\).

Step 3: Write the equation

Using the standard form \((x - h)^2 = 4p(y - k)\) with \(h = 0\), \(k = 6\) and \(p=-3\):
\[
x^{2}=4\times(-3)(y - 6)
\]
\[
x^{2}=-12(y - 6)
\]
\[
x^{2}=-12y+72
\]
\[
x^{2}+12y - 72 = 0
\]
Or in the form \(x^{2}=-12(y - 6)\)

Answer:

The equation of the parabola is \(x^{2}=-12(y - 6)\) (or \(x^{2}+12y - 72 = 0\))

Problem 10:

Given the equation \((y - 6)^{2}=-8(x - 6)\)

Part (a): Coordinates of vertex and focus

Step 1: Identify the vertex

The standard form of a parabola that opens left or right is \((y - k)^{2}=4p(x - h)\), where \((h, k)\) is the vertex. Comparing \((y - 6)^{2}=-8(x - 6)\) with \((y - k)^{2}=4p(x - h)\), we have \(h = 6\), \(k = 6\). So the vertex is \((h,k)=(6, 6)\).

Step 2: Find the value of \(p\)

From the equation \((y - 6)^{2}=-8(x - 6)\), we have \(4p=-8\), so \(p=\frac{-8}{4}=-2\).

Step 3: Find the focus

For a parabola of the form \((y - k)^{2}=4p(x - h)\) that opens left (since \(p<0\)), the focus is at \((h + p, k)\). Substituting \(h = 6\), \(p=-2\) and \(k = 6\), we get the focus as \((6-2,6)=(4, 6)\).

Part (b): Equation of the axis of symmetry and directrix

Step 1: Axis of symmetry

For a parabola of the form \((y - k)^{2}=4p(x - h)\), the axis of symmetry is the horizontal line \(y = k\). Here \(k = 6\), so the axis of symmetry is \(y = 6\).

Step 2: Directrix

The directrix of a parabola of the form \((y - k)^{2}=4p(x - h)\) is the vertical line \(x=h - p\). We know \(h = 6\) and \(p=-2\), so \(x=6-(-2)=6 + 2=8\). So the directrix is \(x = 8\).

Part (c): Direction of opening

Since the coefficient of \((x - 6)\) is negative (\(-8\)) and the parabola is of the form \((y - k)^{2}=4p(x - h)\) with \(p=-2<0\), the parabola opens to the left.

Part (d): Length of the latus rectum

The length of the latus rectum of a parabola of the form \((y - k)^{2}=4p(x - h)\) is \(|4p|\). Here \(4p=-8\), so the length of the latus rectum is \(|-8| = 8\).

Problem 11:

Given the equation \(y=\frac{1}{2}x^{2}+12x - 8\)

Part (a): Coordinates of vertex and focus

Step 1: Complete the square to find the vertex

First, factor out the coefficient of \(x^{2}\) from the \(x\)-terms:
\[
y=\frac{1}{2}(x^{2}+24x)-8
\]
To complete the square inside the parentheses, take half of \(24\) (which is \(12\)) and square it (\(12^{2}=144\)):
\[
y=\frac{1}{2}(x^{2}+24x + 144-144)-8
\]
\[
y=\frac{1}{2}((x + 12)^{2}-144)-8
\]
\[
y=\frac{1}{2}(x + 12)^{2}-72-8
\]
\[
y=\frac{1}{2}(x + 12)^{2}-80
\]
The vertex form of a parabola that opens up or down is \(y=a(x - h)^{2}+k\), where \((h, k)\) is the vertex. Here \(h=-12\), \(k = - 80\), so the vertex is \((-12,-80)\).

Step 2: Find the value of \(p\)

For a parabola of the form \(y=a(x - h)^{2}+k\), the standard form is \((x - h)^{2}=4p(y - k)\), and \(a=\frac{1}{4p}\). From \(y=\frac{1}{2}(x + 12)^{2}-80\), we can rewrite it as \((x + 12)^{2}=2(y + 80)\). Comparing with \((x - h)^{2}=4p(y - k)\), we have \(4p = 2\), so \(p=\frac{2}{4}=\frac{1}{2}\).

Step 3: Find the focus

The focus of a parabola that opens up (since \(a=\frac{1}{2}>0\)) is at \((h,k + p)\). Here \(h=-12\), \(k=-80\) and \(p=\frac{1}{2}\), so the focus is \((-12,-80+\frac{1}{2})=(-12,-\frac{159}{2})\) or \((-12,-79.5)\).

Part (b): Equation of the axis of symmetry and directrix

Step 1: Axis of symmetry

For a parabola of the form \(y=a(x - h)^{2}+k\), the axis of symmetry is the vertical line \(x = h\). Here \(h=-12\), so the axis of symmetry is \(x=-12\).

Step 2: Directrix

The directrix of a parabola of the form \(y=a(x - h)^{2}+k\) is the horizontal line \(y=k - p\). Here \(k=-80\) and \(p=\frac{1}{2}\), so \(y=-80-\frac{1}{2}=-\frac{161}{2}\) or \(y=-80.5\).

Part (c): Direction of opening

Since \(a=\frac{1}{2}>0\) (the coefficient of \((x + 12)^{2}\) is positive), the parabola opens upward.

Part (d): Length of the latus rectum

The length of the latus rectum of a parabola of the form \(y=a(x - h)^{2}+k\) is \(|4p|\). Since \(4p = 2\), the length of the latus rectum is \(|2| = 2\).

Final Answers:

Problem 9:

The equation of the parabola is \(\boldsymbol{x^{2}=-12(y - 6)}\) (or \(x^{2}+12y - 72 = 0\))

Problem 10:

a) Vertex: \(\boldsymbol{(6, 6)}\), Focus: \(\boldsymbol{(4, 6)}\)
b) Axis of symmetry: \(\boldsymbol{y = 6}\), Directrix: \(\boldsymbol{x = 8}\)
c) Direction of opening: \(\boldsymbol{\text{Left}}\)
d) Length of latus rectum: \(\boldsymbol{8}\)

Problem 11:

a) Vertex: \(\boldsymbol{(-12,-80)}\), Focus: \(\boldsymbol{(-12,-\frac{159}{2})}\) (or \((-12,-79.5)\))
b) Axis of symmetry: \(\boldsymbol{x=-12}\), Directrix: \(\boldsymbol{y = -\frac{161}{2}}\) (or \(y=-80.5\))
c) Direction of opening: \(\boldsymbol{\text{Upward}}\)
d) Length of latus rectum: \(\boldsymbol{2}\)