Find Focus of Parabola: y:-(1/4)x^2 + 2x - 5

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Homework Statement



Find the focus of parabola: y:-(1/4)x^2 + 2x - 5

The Attempt at a Solution



Multiplying by 4 to get rid of fraction I get:
4y:-x^2 + 2x - 5

I will bring over the -5:
4y+5:-x^2 +2x

Completing Square I obtain:

4y+6:-(x-1)^2


Is all this correct thus far?
 
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Do you know what a focus of a parabola is? I think it has something to do with polar coordinates.
 
The focus is the central point of a parabola I believe. It determines if parabola points up or down
 
bengaltiger14 said:
The focus is the central point of a parabola I believe. It determines if parabola points up or down
That is the extreme point of a parabola, not a focus.

A physical definition for focus could be this: If you shine parallel beams of light on the parabolic surface, it all reflects and focuses in a special point, called a focus. (A parabolic telescope).

A mathematical one: every point of a parabola has equal distances to a line and a point. The point is called a focus.

Further, if you write down the equation of parabola in polar coordinates in its simplest form, the pole is the focus of parabola.
 
Irid said:
That is the extreme point of a parabola, not a focus.

A physical definition for focus could be this: If you shine parallel beams of light on the parabolic surface, it all reflects and focuses in a special point, called a focus. (A parabolic telescope).

A mathematical one: every point of a parabola has equal distances to a line and a point. The point is called a focus.

Further, if you write down the equation of parabola in polar coordinates in its simplest form, the pole is the focus of parabola.

All very true. But you don't NEED polar coordinates. If you write a standard form for the parabola in cartesian coordinates x^2=4ay the vertex is at the origin and the distance from vertex to focus is 'a'. You just have to write your quadratic in a form similar to the standard form. Like (x-x0)^2=4a(y-y0). Now you have vertex at (x0,y0) and distance to focus 'a'.