Question #7c39a

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Answer 1 · 2018-02-11T03:10:42

$(1, 1.5)$ $(1,1.5)$

Let's rewrite this into form $4 p (y - k) = {(x - h)}^{2}$ $4p(y-k)=(x-h)^2$

$y = \frac{1}{2} {(x - 1)}^{2} + 1$ $y=1/2(x-1)^2+1$

=> $y - 1 = \frac{1}{2} {(x - 1)}^{2}$ $y-1=1/2(x-1)^2$

=> $2 (y - 1) = {(x - 1)}^{2}$ $2(y-1)=(x-1)^2$

We now compare this to $4 p (y - k) = {(x - h)}^{2}$ $4p(y-k)=(x-h)^2$

$2 (y - 1) = {(x - 1)}^{2}$ $2(y-1)=(x-1)^2$
$4 p (y - k) = {(x - h)}^{2}$ $4p(y-k)=(x-h)^2$

It seems to us that:
$h = 1$ $h=1$
$k = 1$ $k=1$
$4 p = 2$ $4p=2$

Let's solve for $p$ $p$ .

$4 p = 2$ $4p=2$

=> $p = \frac{1}{2}$ $p=1/2$

Now, since $x - h$ $x-h$ is being squared, this parabola opens vertically and therefore we deal with $k$ $k$ for the focus.

The focus is $(h, k + p)$ $(h,k+p)$ for any horizontal parabola.

Therefore, the focus is:
$(1, 1 + \frac{1}{2})$ $(1,1+1/2)$

=> $(1, 1.5)$ $(1,1.5)$

That is our answer!