Heres a tricky one my picture wont upload but I need help fi

Here\'s a tricky one... my picture won\'t upload but I need help figuring out how to find an equation of the parabola on the graph.

The vertex of the parabola is at (0,0) and is opening to the right.

There is a square drawn to the left of the y axis, above the x-axis, touching the vertex with an arrow saying \"square has area 16\". The next line to the left is the directrix.

The graph has no numbers listed or even grid lines to determine what the numbers are. Any help you\'re able to provide would be appreciated!

Solution

Alright, here\'s what i understand...

Vertex = (0,0)
and oprning right

So, its gotta be of the form :
y^2 = 4ax

Square has area = 16,
as in side length is 4

So, i am guessing the square side length is going from x = 0
to x = -4, that counts for the 4 units

The next line on the left would make it x = -5

So, the directrix is the line x = -5

And we know that the vertex is given by
x = -p

So, equating, we have :
-p = -5

p = 5

So, we have :
y^2 = 4px

y^2 = 4(5)x

y^2 = 20x ----> ANSWER