Parabolic Equations Using Vertex & Focus

[TheLastLetter]

Homework Statement

Write the equation of the parabola described.
Vertex: (2, 4) Focus: (2,6)

Homework Equations

(x-h)^2 = 4p(y-k)

The Attempt at a Solution

(x-2)^2 = 4(2)(y-4)
x^2-4x+4 = 8y -32

Do I need to isolate one of the variables, or can I leave the equation like this?

 

[whybother]

I'd at least group your constants, to make it look nicer. After you've done that, you could divide through by 8, but it isn't strictly required (unless your teacher said so). Afterall, an equation for a circle is $$x^2 + y^2 = 1$$, and that doesn't have an 'isolated' variable.

 

[Architeuthis Dux]

I would like to clarify that the parabolic equation you have written is not complete. In order to fully describe a parabola, we need to include the value of "p" which represents the distance between the vertex and the focus. This value is crucial in determining the shape and position of the parabola.

To find the value of "p" in this case, we can use the distance formula between the vertex and the focus, which is given by √(x2-x1)^2 + (y2-y1)^2. Plugging in the coordinates of the vertex and focus, we get √(2-2)^2 + (6-4)^2 = √0+4 = 2. Therefore, p=2.

Now we can plug in the value of p into your equation to get the final form: (x-2)^2 = 8(y-4). This equation represents a parabola with the given vertex and focus. If you would like to isolate one of the variables, you can rearrange the equation to get y = (1/8)(x-2)^2 + 4. This equation shows the relationship between x and y, with the vertex (2,4) being the minimum point on the parabola.

I hope this helps clarify your understanding of parabolic equations using vertex and focus. Keep up the good work!

 

[Architeuthis Dux]

Your solution is correct. You can leave the equation as is, but if you want to isolate one of the variables, you can divide both sides by 8 to get the equation in standard form: y = 1/8x^2 - 1/2x + 5. This form may be more useful if you need to graph the parabola or find other points on the curve.