Question

Find the standard form of the equation of the parabola with the given characteristics. Vertex: (-6, 1); focus: (-6, 0)

Answer 1

Answer:
`(x+6)^2=-4(y-1)`

Explanation:

Given

Vertex of the parabola
`(-6,1)`

Focus of the parabola
`(-6,0)`

As the x coordinate of vertex and focus is same and focus lie below the vertex, therefore the parabola is the type of

`(x+6)^2=-4a(y-1)`

distance between Vertex and focus is 1 unit

`\therefore a=1`

Parabola becomes

`\Rightarrow (x+6)^2=-4(y-1)`