Answer:
[tex]\textsf{1.} \quad (x-1)^2=12(y+2)[/tex]
See attachment 1.
2. See attachment 2.
Step-by-step explanation:
Question 1
Given values:
- Vertex: (1, -2)
- Focus: (1, 1)
As the x-value of the vertex and focus is the same, the parabola has a vertical axis of symmetry.
The focus is always on the inside of the parabola.
Since p represents the distance from the vertex to the focus, and the distance from the vertex to the focus is 1 - (-2) = 3, then p = 3.
If p > 0, the parabola opens upwards, and if p < 0, the parabola opens downwards. Therefore, as p > 0, the parabola opens upwards.
[tex]\boxed{\begin{minipage}{5.8 cm}\underline{Standard form of a parabola}\\(with a vertical axis of symmetry)\\\\$(x-h)^2=4p(y-k)$\\\\where:\\ \phantom{ww}$\bullet$ $p\neq 0$\\ \phantom{ww}$\bullet$ Vertex: \;$(h,k)$\\ \phantom{ww}$\bullet$ Focus:\; $(h,k+p)$\\ \phantom{ww}$\bullet$ Directrix: \; $y=(k-p)$\\ \phantom{ww}$\bullet$ Axis of symmetry: \; $x=h$\\\end{minipage}}[/tex]
Therefore:
Substitute the values into the formula:
[tex]\implies (x-h)^2=4p(y-k)[/tex]
[tex]\implies (x-1)^2=4\cdot 3(y-(-2))[/tex]
[tex]\implies (x-1)^2=12(y+2)[/tex]
See attachment 1 for the graph of the parabola.
Question 2
Given equation of a parabola:
[tex](y-4)^2=8(x+2)[/tex]
As the y-variable is contained within the squared part of the equation, the parabola has a horizontal axis of symmetry.
[tex]\boxed{\begin{minipage}{5.8 cm}\underline{Standard form of a parabola}\\(with a horizontal axis of symmetry)\\\\$(y-k)^2=4p(x-h)$\\\\where:\\ \phantom{ww}$\bullet$ $p\neq 0$\\ \phantom{ww}$\bullet$ Vertex: \;$(h,k)$\\ \phantom{ww}$\bullet$ Focus:\; $(h+p,k)$\\ \phantom{ww}$\bullet$ Directrix: \; $x=(h-p)$\\ \phantom{ww}$\bullet$ Axis of symmetry: \; $y=k$\\\end{minipage}}[/tex]
Therefore:
- h = -2
- k = 4
- 4p = 8 ⇒ p = 2
If p > 0, the parabola opens to the right, and if p < 0, the parabola opens to the left. Therefore, as p > 0, the parabola opens to the right.
Find the y-intercepts of the graph by substituting x = 0 into the equation:
[tex]\implies (y-4)^2=8(0+2)[/tex]
[tex]\implies (y-4)^2=16[/tex]
[tex]\implies y-4=\pm4[/tex]
[tex]\implies y=4\pm4[/tex]
[tex]\implies y=0, y=8[/tex]
To sketch the graph of the parabola, plot:
- Vertex = (-2, 4)
- Focus = (0, 4)
- Axis of symmetry: y = 4
- y-intercepts: (0, 0) and (0, 8)
Draw a curve through the vertex and y-intercepts, opening to the right.
Use the axis of symmetry to ensure the curve is symmetric.
See attachment 2 for the graph of the parabola.
