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Parabola standard form of equation at x = -5

Joystar1977

Active member
Jul 24, 2013
119
Find the standard form of the equation of the parabola with the given characteristics and vertex at the origin. Passes through the point (-5, 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
Jul 24, 2013
119
re: Parabola standard form of equation at x=-5

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 (-5, 1/8); vertical axis. Would this problem be correct if I work it out this way?

The axis is vertical so I know that the vertex is (0,0).

y = a (x - h)^2 + k

Since, I know that the vertex is at (0,0), then I know that h = k = 0 and thought that this would be the proper way to work the problem out.

y = ax ^2

1/8 = 25a

a = 1/200

so, y = 1/200 x ^2

Is this correct way to work out this problem? If not, then can somebody please help me? I am a little confused on which formula to use for this problem.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Parabola standard form of equation at x=-5

Personally, I would just look at the fact that the vertex is at the origin, which means, given the vertical axis of symmetry, that the parabola must be of the form:

\(\displaystyle y=ax^2\)

Now, using the other given point, can you determine the value of the parameter $a$?
 

Joystar1977

Active member
Jul 24, 2013
119
Re: Parabola standard form of equation at x=-5

What do you mean by parameter? Isn't the value of a = 1/200?

Personally, I would just look at the fact that the vertex is at the origin, which means, given the vertical axis of symmetry, that the parabola must be of the form:

\(\displaystyle y=ax^2\)

Now, using the other given point, can you determine the value of the parameter $a$?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Parabola standard form of equation at x=-5

What do you mean by parameter? Isn't the value of a = 1/200?
In this context, you can think of a parameter as a constant whose value we must determine. We are told the parabola passes through the point \(\displaystyle (x,y)=\left(-5,\frac{1}{8} \right)\). And so you are right, we find:

\(\displaystyle \frac{1}{8}=a(-5)^2=25a\implies a=\frac{1}{200}\)

And so we know the parabola satisfying the given conditions is:

\(\displaystyle y=\frac{x^2}{200}\)