Question

Write the equation of the conic section below.

Answer 1

`(x-3)^2=8(y+1)`

Explanation:

Standard form of a parabola with a vertical axis of symmetry:

`(x-h)^2=4p(y-k) \quad \textsf{where}\:p\\eq 0`

`\textsf{Vertex}=(h, k)`

`\textsf{Focus}=(h,k+p)`

`\textsf{Directrix}:y=(k-p)`

`\textsf{Axis of symmetry}:h=k`

If p > 0, the parabola opens upwards, and if p < 0, the parabola opens downwards.

From inspection of the graph:

• Vertex = (3, -1)
• Directrix = y = -3

Therefore:

• h = 3
• k = -1
• k - p = -3

Use the Directrix equation to find p

⇒ y = (k - p)

⇒ -3 = -1 - p

⇒ p = 2

Therefore the equation of the conic section is:

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

`\implies (x-3)^2=8(y+1)`

Rearranging in standard form
`ax^2+bx+c`:

`\implies (x-3)^2=8y+8`

`\implies x^2-6x+9-8=8y`

`\implies x^2-6x+1=8y`

`\implies y=(1)/(8)x^2-(3)/(4)x+(1)/(8)`