Question

Part 3. Identify the vertex, focus, and directrix of each. Then sketch the graph.

1. x = -1/2(y -5)^2 - 1

2. y = 1/3(x + 4)^2 -1

Answer 1

Answer:
`\bold{1.\quad \text{Vertex}=(-1,5)\qquad \text{Focus}=\bigg(-(3)/(2),5\bigg)\qquad \text{Directrix}:x=-(1)/(2)}`

`\bold{2.\quad \text{Vertex}=(-4,-1)\quad \text{Focus}=\bigg(-4,-(1)/(4)\bigg)\qquad \text{Directrix}:y=-(7)/(4)}`

Explanation:

The vertex form of a parabola is x = a(y - k)² + h or y = a(x - h)² + k

• (h, k) is the vertex

`\bullet\quad a=(1)/(4p)`

• p is the distance from the vertex to the focus
• -p is the distance from the vertex to the directrix

1)

`x=-(1)/(2)(y-5)^2-1\qquad \rightarrow \qquad a=-(1)/(2)\quad (h,k)=(-1,5)\\\\a=(1)/(4p)\qquad \rightarrow \quad -(1)/(2)=(1)/(4p)\qquad \rightarrow \quad-2=4p\qquad \rightarrow \quad-(1)/(2)=p\\\\\\\text{Focus = Vertex + p}\\\\.\qquad =(-2)/(2)+(-1)/(2)\\\\.\qquad =-(3)/(2)\qquad \rightarrow \qquad \text{Focus}=\bigg(-(3)/(2),5\bigg)\\`

`\text{Directrix: x = Vertex - p}\\\\.\qquad \qquad x=(-2)/(2)-(-1)/(2)\\\\.\qquad \qquad x=-(1)/(2)`

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2)

`y=(1)/(3)(x+4)^2-1\qquad \rightarrow \qquad a=\frac{1}{}\quad (h,k)=(-4,-1)\\\\a=(1)/(4p)\qquad \rightarrow \quad (1)/(3)=(1)/(4p)\qquad \rightarrow \quad3=4p\qquad \rightarrow \quad(3)/(4)=p\\\\\\\text{Focus = Vertex + p}\\\\.\qquad =(-4)/(4)+(3)/(4)\\\\.\qquad =-(1)/(4)\qquad \rightarrow \qquad \text{Focus}=\bigg(-4,-(1)/(4)\bigg)\\`

`\text{Directrix: y = Vertex - p}\\\\.\qquad \qquad y=(-4)/(4)-(3)/(4)\\\\.\qquad \qquad y=-(7)/(4)`