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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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