The equation of a parabola is given. y=1/8x^2+4x+20 What are the coordinates of the focus of the parabola?

Guest Jun 4, 2017

#1**0 **

The parabola focus is the point wherein the distance to a point on a parabola is equidistant to the distance to the directrix!

To find the focus, convert the quadratic to vertex form, \(y=a(x-h)^2+k\) where \((h,k+\frac{1}{4a})\) is the focus. Let's try and do this:

\(y=\frac{1}{8}x^2+4x+20\) | This is the original quadratic equation. In order ro convert the quadratic to the desired form above, we need to use a method called "completing the square." First, subtract 20 on both sides. |

\(y-20=\frac{1}{8}x^2+4x\) | Multiply by 8 on both sides to get rid of the pesky fraction |

\(8y-160=x^2+32x\) | This is where completing the square comes in handy. Do the linear x-term and half it. Take that quantity and square it. Add it to both sides. |

\(8y-160+(\frac{32}{2})^2=x^2+32x+(\frac{32}{2})^2\) | Simplify both sides of the equation |

\(8y+96=x^2+32x+256\) | What's the point of doing all this work? Well, the right hand side is a perfect square trinomial. |

\(8y+96=(x+16)^2\) | Subtract 96 on both sides of the equation |

\(8y=(x+16)^2-96\) | Divide by 8 on both sides |

\(y=\frac{1}{8}(x+16)^2-12\) | |

Our quadratic equation is finally in vertex form. Now, we can find the focus by using the formula I mentioned above, \((h,k+\frac{1}{4a})\). Let's plug those values into this quadratic equation. First, identify what *h, k, and a *are.

h=-16

k=-12

a=1/8

Let's plug these values in:

\((-16,-12+\frac{1}{4(\frac{1}{8})})\) | Do 4*1/8 first. |

\((-16,-12+\frac{1}{\frac{1}{2}})\) | I'll use a fraction rule that states that \(\frac{a}{\frac{b}{c}}=\frac{ac}{b}\) |

\((-16,-12+2)\) | Continue simplifying. |

\((-16,-10)\) | |

Now, you are finally done. The point of the focus is \((-16,-10)\).

TheXSquaredFactor
Jun 5, 2017