Parabola standard form of equation at x = -1

Joystar1977

Active member
Find the standard form of the equation of the parabola with the given characteristics and vertex at the origin. Passes through the point (-1, 1/8); vertical axis.

I know that there is no focus of the parabola or equation given for this problem, so how would I solve this problem? Is the correct formula to use the following?:

x^2= 4py

Are these the correct steps to take?

1. Write original equation

2. Divide each side by number given.

3. Write in standard form.

Joystar1977

Active member
re: Parabola standard form of equation at x=-1

I have two different formulas for using the conic section of a parabola, can someone please tell me which is correct for this type of problem?

The first one is as follows:
Type: Parabola
General Equation: y = a (x-h)^2 + k
Standard Form: (x - h) ^2 = 4p (y-k)

Notation:
1. x2 term and y1 term.
2. (h,k) is vertex.
3. (h, k does not equal p) is center of focus, where p = 1/4a.
4. y =k does not equal p is directrix equation, where p = 1/4a.

Value:
1. a >0, then opens up.
2. a < 0, then opens down.
3. x = h is equation of line of symmetry.
4. Larger [a] = thinner parabola; smaller [a] = fatter parabola.

Type: Parabola
General Equation: x = a (y-k)^2 + h
Standard Form: (y-k)^2 = 4p(x-h)

Notation:
1. x1 term and y2 term.
2. (h,k) is vertex.
3. (h does not equal p, k) is focus, where p = 1/4a.
4. x = h does not equal p is directrix equation, where p = 1/4a.

Values:

1. a > 0, then opens right.
2. a < 0, then opens left.
3. y = k is equation of line of symmetry.

In this problem, find the standard form of the equation of the parabola with the given characteristics and vertex at the origin. Passes through the point (-1, 1/8); vertical axis. Would this problem be correct if I work it out this way?

Vertex is at (0,0), then is the equation y = ax^2?

1/8 = a(-1)^2

a = 1/8

y = (1/8)x^2

STANDARD FORM:

y = (1/8) x^2

8y = x^2

x^2 - 8y = 0

Is this the correct way to solve this problem? Which formula am I suppose to use? If this is not right, then can somebody please help me with this?

HallsofIvy

Well-known member
MHB Math Helper
Re: Parabola standard form of equation at x=-1

Except for the fact that you have the parabola going through a point with x= -1 instead of x= -5, this is exactly the same as your post "Parabola standard form of equation at x= -4" and every answer there applies to this.

You ask "which of these formulas is correct for this type of problem
General Equation: y = a (x-h)^2 + k
Standard Form: (x - h) ^2 = 4p (y-k)"

The answer is that they are the same formula except that the "a" in one is "1/4p" where p is from the other.

In both problems, the "vertex is at the origin" so h= k= 0 and the two equations become
General Equation: y = ax^2
Standard Form: x^2 = 4py
and, again, they are the same with a= 1/4p. Here, you are told that the graph goes through the point (-1, 1/8) so put x= -1, y= 1/8 into either and solve for a or p, as appropriate.