Question

What are the coordinates of the focus of the parabola?y=18x2+2x

Answer 1

The equation of the given parabola is

`y=(1)/(8)x^2_{}+2x`

Rewrite the equation in the vertex form

`y=a(x-h)^2+k`

The equation becomes

`\begin{gathered} y=(1)/(8)x^2+2x \\ y=(1)/(8)(x^2+16x) \\ 8y=x^2+16x \\ 8y=x^2+6x+64-64 \\ 8y=(x+8)^2-64 \end{gathered}`

Divide through the equation by 8

This gives

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

Comparing the equation with the vertex form

It follows

`a=(1)/(8),h=-8,k=-8`

The focus of a parabola in vertex form is given as

`F=(h,k+(1)/(4a))`

Substitute h = -8, k = -8 and a = 1/8 into the formula for focus

This gives

`F=(-8,-8+(1)/(4((1)/(8))))`

Simplify the expression

`\begin{gathered} F=(-8,-8+(1)/((1)/(2))) \\ F=(-8,-8+2) \\ F=(-8,-6) \end{gathered}`

Therefore, the focus of the parabola is at (-8, -6)