Answer :
Sure! Let’s go through each equation step-by-step to determine each of the requested properties of the parabolas.
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### Equation a ###
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### Given equation:
[tex]\[ x = \frac{1}{12} y^2 \][/tex]
A. If the parabola opens up, down, left, or right:
The given equation is in the form [tex]\( x = a(y - k)^2 + h \)[/tex]. Here, [tex]\( a = \frac{1}{12} \)[/tex], and since [tex]\( a \)[/tex] is positive, the parabola opens to the right.
B. The vertex of the parabola:
For the equation [tex]\( x = \frac{1}{12} y^2 \)[/tex], the vertex [tex]\( (h, k) \)[/tex] is [tex]\((0, 0)\)[/tex].
C. The location of the focus:
The focus of a parabola [tex]\( x = a(y - k)^2 + h \)[/tex] is located at [tex]\( \left(h + \frac{1}{4a}, k\right) \)[/tex]. Here [tex]\( a = \frac{1}{12} \)[/tex], so the focus is:
[tex]\[ \left(0 + \frac{1}{4 \times \frac{1}{12}}, 0\right) = \left(3, 0\right) \][/tex]
D. The equation of the directrix:
The directrix of a parabola [tex]\( x = a(y - k)^2 + h \)[/tex] has the equation [tex]\( x = h - \frac{1}{4a} \)[/tex]. Here [tex]\( a = \frac{1}{12} \)[/tex], so the equation of the directrix is:
[tex]\[ x = 0 - \frac{1}{4 \times \frac{1}{12}} = -3 \][/tex]
E. The equation of the axis of symmetry:
The axis of symmetry for the parabola [tex]\( x = a(y - k)^2 + h \)[/tex] is the line [tex]\( y = k \)[/tex]. Here [tex]\( k = 0 \)[/tex], so the equation of the axis of symmetry is:
[tex]\[ y = 0 \][/tex]
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### Equation b ###
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### Given equation:
[tex]\[ y - 2 = -\frac{1}{16} (x - 1)^2 \][/tex]
A. If the parabola opens up, down, left, or right:
The given equation is in the form [tex]\( y - k = a(x - h)^2 \)[/tex]. Here, [tex]\( a = -\frac{1}{16} \)[/tex], and since [tex]\( a \)[/tex] is negative, the parabola opens downwards.
B. The vertex of the parabola:
For the equation [tex]\( y - 2 = -\frac{1}{16}(x - 1)^2 \)[/tex], the vertex [tex]\( (h, k) \)[/tex] is [tex]\((1, 2)\)[/tex].
C. The location of the focus:
The focus of a parabola [tex]\( y - k = a(x - h)^2 \)[/tex] is located at [tex]\( \left(h, k + \frac{1}{4a}\right) \)[/tex]. Here [tex]\( a = -\frac{1}{16} \)[/tex], so the focus is:
[tex]\[ \left(1, 2 + \frac{1}{4 \times -\frac{1}{16}} \right) = \left(1, 6\right) \][/tex]
D. The equation of the directrix:
The directrix of a parabola [tex]\( y - k = a(x - h)^2 \)[/tex] has the equation [tex]\( y = k - \frac{1}{4a} \)[/tex]. Here [tex]\( a = -\frac{1}{16} \)[/tex], so the equation of the directrix is:
[tex]\[ y = 2 - \frac{1}{4 \times -\frac{1}{16}} = -2 \][/tex]
E. The equation of the axis of symmetry:
The axis of symmetry for the parabola [tex]\( y - k = a(x - h)^2 \)[/tex] is the line [tex]\( x = h \)[/tex]. Here [tex]\( h = 1 \)[/tex], so the equation of the axis of symmetry is:
[tex]\[ x = 1 \][/tex]
### Summary:
For equation [tex]\( a \)[/tex]:
A. Opens right
B. Vertex at [tex]\((0, 0)\)[/tex]
C. Focus at [tex]\((3, 0)\)[/tex]
D. Directrix: [tex]\(x = -3\)[/tex]
E. Axis of symmetry: [tex]\(y = 0\)[/tex]
For equation [tex]\( b \)[/tex]:
A. Opens down
B. Vertex at [tex]\((1, 2)\)[/tex]
C. Focus at [tex]\((1, 6)\)[/tex]
D. Directrix: [tex]\(y = -2\)[/tex]
E. Axis of symmetry: [tex]\(x = 1\)[/tex]
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### Equation a ###
##################
### Given equation:
[tex]\[ x = \frac{1}{12} y^2 \][/tex]
A. If the parabola opens up, down, left, or right:
The given equation is in the form [tex]\( x = a(y - k)^2 + h \)[/tex]. Here, [tex]\( a = \frac{1}{12} \)[/tex], and since [tex]\( a \)[/tex] is positive, the parabola opens to the right.
B. The vertex of the parabola:
For the equation [tex]\( x = \frac{1}{12} y^2 \)[/tex], the vertex [tex]\( (h, k) \)[/tex] is [tex]\((0, 0)\)[/tex].
C. The location of the focus:
The focus of a parabola [tex]\( x = a(y - k)^2 + h \)[/tex] is located at [tex]\( \left(h + \frac{1}{4a}, k\right) \)[/tex]. Here [tex]\( a = \frac{1}{12} \)[/tex], so the focus is:
[tex]\[ \left(0 + \frac{1}{4 \times \frac{1}{12}}, 0\right) = \left(3, 0\right) \][/tex]
D. The equation of the directrix:
The directrix of a parabola [tex]\( x = a(y - k)^2 + h \)[/tex] has the equation [tex]\( x = h - \frac{1}{4a} \)[/tex]. Here [tex]\( a = \frac{1}{12} \)[/tex], so the equation of the directrix is:
[tex]\[ x = 0 - \frac{1}{4 \times \frac{1}{12}} = -3 \][/tex]
E. The equation of the axis of symmetry:
The axis of symmetry for the parabola [tex]\( x = a(y - k)^2 + h \)[/tex] is the line [tex]\( y = k \)[/tex]. Here [tex]\( k = 0 \)[/tex], so the equation of the axis of symmetry is:
[tex]\[ y = 0 \][/tex]
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### Equation b ###
##################
### Given equation:
[tex]\[ y - 2 = -\frac{1}{16} (x - 1)^2 \][/tex]
A. If the parabola opens up, down, left, or right:
The given equation is in the form [tex]\( y - k = a(x - h)^2 \)[/tex]. Here, [tex]\( a = -\frac{1}{16} \)[/tex], and since [tex]\( a \)[/tex] is negative, the parabola opens downwards.
B. The vertex of the parabola:
For the equation [tex]\( y - 2 = -\frac{1}{16}(x - 1)^2 \)[/tex], the vertex [tex]\( (h, k) \)[/tex] is [tex]\((1, 2)\)[/tex].
C. The location of the focus:
The focus of a parabola [tex]\( y - k = a(x - h)^2 \)[/tex] is located at [tex]\( \left(h, k + \frac{1}{4a}\right) \)[/tex]. Here [tex]\( a = -\frac{1}{16} \)[/tex], so the focus is:
[tex]\[ \left(1, 2 + \frac{1}{4 \times -\frac{1}{16}} \right) = \left(1, 6\right) \][/tex]
D. The equation of the directrix:
The directrix of a parabola [tex]\( y - k = a(x - h)^2 \)[/tex] has the equation [tex]\( y = k - \frac{1}{4a} \)[/tex]. Here [tex]\( a = -\frac{1}{16} \)[/tex], so the equation of the directrix is:
[tex]\[ y = 2 - \frac{1}{4 \times -\frac{1}{16}} = -2 \][/tex]
E. The equation of the axis of symmetry:
The axis of symmetry for the parabola [tex]\( y - k = a(x - h)^2 \)[/tex] is the line [tex]\( x = h \)[/tex]. Here [tex]\( h = 1 \)[/tex], so the equation of the axis of symmetry is:
[tex]\[ x = 1 \][/tex]
### Summary:
For equation [tex]\( a \)[/tex]:
A. Opens right
B. Vertex at [tex]\((0, 0)\)[/tex]
C. Focus at [tex]\((3, 0)\)[/tex]
D. Directrix: [tex]\(x = -3\)[/tex]
E. Axis of symmetry: [tex]\(y = 0\)[/tex]
For equation [tex]\( b \)[/tex]:
A. Opens down
B. Vertex at [tex]\((1, 2)\)[/tex]
C. Focus at [tex]\((1, 6)\)[/tex]
D. Directrix: [tex]\(y = -2\)[/tex]
E. Axis of symmetry: [tex]\(x = 1\)[/tex]
