Question

Hello! Can someone please help me. I am in trouble, hope ypu help me guys, with complete answer! THANK YOU​

Answer 1

`\textsf{1.} \quad (x-1)^2=12(y+2)`

See attachment 1.

2. See attachment 2.

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.

`\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$ $p\\eq 0$\\ \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}}`

Therefore:

• h = 1
• k = -2
• p = 3

Substitute the values into the formula:

`\implies (x-h)^2=4p(y-k)`

`\implies (x-1)^2=4\cdot 3(y-(-2))`

`\implies (x-1)^2=12(y+2)`

See attachment 1 for the graph of the parabola.

Question 2

Given equation of a parabola:

`(y-4)^2=8(x+2)`

As the y-variable is contained within the squared part of the equation, the parabola has a horizontal axis of symmetry.

`\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$ $p\\eq 0$\\ \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}}`

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:

`\implies (y-4)^2=8(0+2)`

`\implies (y-4)^2=16`

`\implies y-4=\pm4`

`\implies y=4\pm4`

`\implies y=0, y=8`

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.