Question

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

Answer 1

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

Step-by-step 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

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