What is the focal diameter of: y = − 1 8 x2

Answer 1

y = (-1/8)x^2 is a downward opening parabola with vertex = the origin (0,0)

relatively flat.

focus and directrix are equidistant from the vertex. focus has an x coordinate = 0 and y coordinate <0

axis of symmetry is the y axis or x=0

find one more point on the parabola, go horizontally from the focus to the parabola, that distance equals the distance from the parabola at that point to the directrix. its x coordinate = -2 times its y coordinate. it helps to graph a rough sketch of the parabola.

y = (-1/8)x^2

y = (-1/8)(-2y)^2

y = -4y^2/8 = -y^2/2

2y = -y^2

2 =- y

x=-2y = -2(-2) = 4

plug that point (4,-2) into y =(-1/8)x^2

-2 = (-1/8)(4)^2 = -16/8 = -2 = the y coordinate of the vertex, x coordinate = 0

the focus = (0,-2)

distance from the focus to the vertex = 2

distance from the directrix to the vertex = 2

the latus rectum = distance from the parabola points horizontally left and right of the focus = from (4,-2) to (-4,-2) = 2(4) = 8. "Latus rectum" is the math term for that distance.

Answer 2

Because y= (-1/8) x² has a negative coefficient on the x² term, we know it opens down. Since there is no horizontal or vertical shift, we know the vertex is at the origin. You should make a very rough sketch of this on your paper. (Just draw an x axis, a y axis, and a parabola opening down from the origin.) Directly under the vertex somewhere is the focus.

The focal diameter of a parabola is the chord, or diameter, that goes through the focus of the parabola. It's a line segment that connects the sides of a parabola and is perpendicular to the axis of symmetry.

The focus and directrix of a parabola are p units away from the vertex, when 4p is the coefficient of the x² term. Set 4p=-1/8 and multiply both sides by -8 to get p= -2.

So the focus is 2 down from the vertex at (0,-2).

Given y= -2, there are two possible x values. Solve for these by plugging y= -2 into the original equation.

-2 = (-1/8)x². Multiply both sides by -8:

16 = x². And take the square root of both sides:

± 4 = x.

Now we have the endpoints of the focal diameter: (-4,-2) and (4,2). The distance between -4 and 4 is 8, which is the length of the focal diameter.

I hope this helps. Quadratics are so cool! Let me know if you have any more questions?

Take care, and remember to have fun!

Liz Z.