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Thread: Parabola standard form of equation

  1. #1

    Parabola standard form of equation

    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.

    Is the correct formula as follows:

    x^2 = 4py

    Are these correct steps to take for the above problem???

    1. Write original equation.

    2. Divide each side by the given number in equation.

    3. Write in standard form.

    What if there is no focus given for this above problem or a focus of the parabola for an equation? How would I then end up solving this problem? Please help because I am lost and confused.

    How do I know which formula to use when it comes to a conic section of the parabola?

    For example, I have two separate things listed in my textbook as follows when it comes to the parabola.

    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.
    Last edited by Joystar77; 10-14-2013 at 07:03 PM.

  2. #2
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Joystar77 View Post
    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.

    Is the correct formula as follows:

    x^2 = 4py
    Ironically, there is no standard definition of "standard form". You'll have to check your book or your class notes for the form that you are expected to use.

    Quote Originally Posted by Joystar77 View Post
    Are these correct steps to take for the above problem???

    1. Write original equation.
    What do you mean by "the original equation"? How do you propose to go from what you've been given (the vertex, another point, and the direction of the axis) to "the original equation"?

    Quote Originally Posted by Joystar77 View Post
    2. Divide each side by the given number in equation.
    I don't know what this means...? Sorry.

    Quote Originally Posted by Joystar77 View Post
    What if there is no focus given for this above problem or a focus of the parabola for an equation? How would I then end up solving this problem?
    Take the information you have been given, and see where it takes you. For instance, do you "need" the focus, given what you already have? If you do, what information have they given you, which might relate the focus to the vertex, etc?

    It's hard to provide specific advice to a general question. If you're needing lessons on working with parabola equations, try Google.

  3. #3

    Parabola standard form of the equation.

    Stapel:

    I am not trying to play around here if you think I am trying to just joke or play around with this problem. Alright I understand what you said that there is no standard definition of "standard form". The information I got was off of an example that I printed off from google.com. Here is the sample of the problem:

    Find the focus of the parabola whose equation is y = -2x^2.

    Solution: Because the squared term in the equation involves x, you know that the axis is vertical, and the equation is of the form

    x^2= 4py.

    You can write the original equation in this form, as shown.

    -2x^2 = y; Write original equation.

    x^2 = -1/2y; Divide each side by -2.

    x^2 = 4 (-1/8)y; Write in standard form.

    So, p = -1/8. Because p is negative, the parabola opens downward and the focus of the parabola is (0,p) = (0, -1/8).

    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.

    How do you solve this problem when there is no equation given as to the focus of the parabola?


    Quote Originally Posted by stapel View Post
    Ironically, there is no standard definition of "standard form". You'll have to check your book or your class notes for the form that you are expected to use.


    What do you mean by "the original equation"? How do you propose to go from what you've been given (the vertex, another point, and the direction of the axis) to "the original equation"?


    I don't know what this means...? Sorry.


    Take the information you have been given, and see where it takes you. For instance, do you "need" the focus, given what you already have? If you do, what information have they given you, which might relate the focus to the vertex, etc?

    It's hard to provide specific advice to a general question. If you're needing lessons on working with parabola equations, try Google.

  4. #4
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    The standard form of the equation of a parabola is [tex]y=ax^2 + bx +c[/tex]

    The vertex form of the equation of a parabola is [tex]y=a(x-h)^2 + k[/tex] where (h,k) is the vertex.

    Step 1: Plug (h,k) into the vertex form. This is equation 1.

    Step 2: Plug the given point (x,y) into eq 1 to solve for a.

    Step 3: Plug a into equation 1. This is now eq 2.

    Step 4: Write eq 2 in standard form.

    Since the vertex is (0, 0), this problem is pretty simple.
    Last edited by fcabanski; 10-14-2013 at 07:01 PM.

  5. #5

    Standard Form of the equation of a parabola

    fcabanski: Please work with me step-by-step so I can understand this! I get totally lost and confused when there is suppose to be one formula and it's like I am being given four different formulas to use for 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.

    Quote Originally Posted by fcabanski View Post
    The standard form of the equation of a parabola is [tex]y=ax^2 + bx +c[/tex]

    The vertex form of the equation of a parabola is [tex]y=a(x-h)^2 + k[/tex] where (h,k) is the vertex.

    Step 1: Plug (h,k) into the vertex form. This is equation 1.

    Step 2: Plug the given point (x,y) into eq 1 to solve for a.

    Step 3: Plug a into equation 1. This is now eq 2.

    Step 4: Write eq 2 in standard form.

    Since the vertex is (0, 0), this problem is pretty simple.

  6. #6
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    Step 1: Plug (h,k) into the vertex form. This is equation 1.

    The vertex is at the origin (0,0). [tex]y = a(x-0)^2 + 0[/tex] which is [tex]y=ax^2[/tex]

    Step 2: Plug the given point (x,y) into eq 1 to solve for a.

    The given point is (-1, 1/8): [tex]1/8 = a(-1)^2 = a = 1/8 [/tex]

    Step 3: Plug a into equation 1. This is now eq 2.

    [tex]y=(1/8)x^2[/tex]

    Step 4: Write eq 2 in standard form.

    That's already done. [tex]y=(1/8)x^2 + 0x + 0[/tex] is [tex]y = (1/8)x^2[/tex]

    Since the vertex is (0, 0), this problem is pretty simple.
    Last edited by fcabanski; 10-14-2013 at 07:24 PM.

  7. #7
    How do you know that the vertex at the origin is (0,0)? How did you make up an equation to become Equation #1? How did you make up an equation to become Equation #2? I don't understand how you came up with the answers to these questions.

    Quote Originally Posted by fcabanski View Post
    Step 1: Plug (h,k) into the vertex form. This is equation 1.

    The vertex is at the origin (0,0). [tex]y = a(x-0)^2 + 0[/tex] which is [tex]y=ax^2[/tex]

    Step 2: Plug the given point (x,y) into eq 1 to solve for a.

    The given point is (-1, 1/8): [tex]1/8 = a(-1)^2 = a = 1/8 [/tex]

    Step 3: Plug a into equation 1. This is now eq 2.

    [tex]y=(1/8)x^2[/tex]

    Step 4: Write eq 2 in standard form.

    That's already done. [tex]y=(1/8)x^2 + 0x + 0[/tex] is [tex]y = (1/8)x^2[/tex]

    Since the vertex is (0, 0), this problem is pretty simple.

  8. #8
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    How do you know that the vertex at the origin is (0,0)?

    The problem stated that the vertex is at the origin. The origin is (0,0). It's a basic fact of the coordinate plane.

    How did you make up an equation to become Equation #1?

    I didn't make up the equation. The vertex form of the parabola equation is valid for every parabola. It is a general equation. It is used when you know the coordinates of the vertex.


    How did you make up an equation to become Equation #2?

    I didn't make it up. I plugged in the given information. Substitute (0,0) for (h,k) in the vertex form of the equation. Solve for a. Plug that back into the general vertex form equation.

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