Question

Determine the equation of the parabola with focus (3,6) and vertex (3,2).

Answer 1

The standard form equation of a parabola is

`\begin{gathered} 4p(y-k)=(x-h)^2 \\ \text{ Where } \\ (h,k)\text{ is the vertex and} \\ p\text{ is the distance between the vertex and the focus} \end{gathered}`

To find the value of p, we can use the distance formula or operate like this:

`p=6-2=4`

Because:

Now, we can find the equation of parabola:

`\begin{gathered} p=4 \\ h=3 \\ k=2 \end{gathered}`
`\begin{gathered} 4p(y-k)=(x-h)^2 \\ 4\cdot4(y-2)=(x-3)^2 \\ 16(y-2)=(x-3)^2 \\ \text{ }16(y-2)=(x-3)(x-3) \\ \text{ Apply the distributive property} \\ 16\cdot y-16\cdot2=x\cdot x-3\cdot x-3\cdot x+9 \\ 16y-32=x^2-6x+9 \\ \text{ Subtract }x^2,\text{ add 6x and subtract 9 from both sides of the equation} \\ 16y-32-x^2+6x-9=x^2-6x+9-x^2+6x-9 \\ 16y-32-x^2+6x-9=0 \\ \text{ Add similar terms} \\ 16y-x^2+6x-41=0 \\ \text{ Reorder} \\ -x^2+6x+16y-41=0 \\ \text{ Multiply by -1 from both sides} \\ -1(-x^2+6x+16y-41)=-1\cdot0 \\ -1\cdot-x^2-1\cdot6x-1\cdot16y-1\cdot-41=0 \\ x^2-6x-16y+41=0 \end{gathered}`

Therefore, the equation of the parabola is

`x^2-6x-16y+41=0`