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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
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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

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