SOLUTION: I am taking integrated algebra online, and my teacher gave us two standard equations for a parabola in the conic section, but she failed to explain how we decide which one to use w

Question 464984: I am taking integrated algebra online, and my teacher gave us two standard equations for a parabola in the conic section, but she failed to explain how we decide which one to use when given a problem like "focus (-2,1)and directrix x = 3". the two equations she gave us are the vertical equation "(x - h)^2 = 4p (y - k)" and the horizontal equation "(y - k)^2 = 4p (x - h) i know how to figure out what h, k, and p are but she didn't explain how we are suppose to determine which equation to plug our answers into. The two equations give you two different answers when you figure the problem out.
Found 2 solutions by solver91311, lwsshak3:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Your given directrix is a vertical line (), that means you should use the form that has the variable squared. If the directrix is horizontal (), then use the form with the variable squared.

John

My calculator said it, I believe it, that settles it

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
I am taking integrated algebra online, and my teacher gave us two standard equations for a parabola in the conic section, but she failed to explain how we decide which one to use when given a problem like "focus (-2,1)and directrix x = 3". the two equations she gave us are the vertical equation "(x - h)^2 = 4p (y - k)" and the horizontal equation "(y - k)^2 = 4p (x - h) i know how to figure out what h, k, and p are but she didn't explain how we are suppose to determine which equation to plug our answers into. The two equations give you two different answers when you figure the problem out.
...
Let me start my explanation by listing the four different configurations you must consider when working with these problems. Also, lets assume the vertices are at the origin, that is, (h,k) is (0,0).
x^2=4py
This is a parabola that opens upwards, with vertex at (0,0) and a vertical axis of symmetry, x=0 or y-axis
..
x^2=-4py
This is a parabola that opens downwards, with vertex at (0,0) and vertical axis of symmetry, x=0 or y-axis
..
y^2=4px
This is a parabola that opens rightwards, with vertex at (0,0) and horizontal axis of symmetry, y=0 or x-axis
..
y^2=-4px
This is a parabola that opens leftwards, with vertex at (0,0) and horizontal axis of symmetry, y=0 or x-axis
...
Moving the vertex of center, you will get the following four corresponding configurations:
(x-h)^2=4p(y-k) with (h,k) being the (x,y) coordinates of the vertex, vertical axis of symmetry, opens upwards.
(x-h)^2=-4p(y-k) with (h,k) being the (x,y) coordinates of the vertex, vertical axis of symmetry, opens downwards
(y-k)^2=4p(x-h) with (h,k) being the (x,y) coordinates of the vertex,horizontal axis of symmetry, opens rightwards
(y-k)^2=-4p(x-y) with (h,k) being the (x,y) coordinates of the vertex,horizontal axis of symmetry, opens leftwards
..
On the axis of symmetry, p=distance from the vertex to focus and in the opposite direction, distance from vertex to directrix. You will usually see 4p as the coefficient of the right term, so you can find p by dividing this coefficient by 4. The focal point is always located inside the curve and the directrix is line located on the other side of the vertex. The directrix must be expressed as an equation like x=0, y=3, etc. Also, 4p=focal width, a vertical or horizontal line line through the focus intersecting the curve on both sides of the axis of symmetry.
Hope this helps!
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