Question

Samara is adjusting a satellite because she finds it is not focusing the incoming radio waves perfectly. The shape of her satellite can be modeled by (y-3)^2=8(x-4)where x and y are modeled in inches. She realizes that the static is a result of the feed antenna shifting slightly off the focus point. What is the focus point of the satellite?

A. (–3, –6)
B. (–3, –4)
C. (3, 6)
D. (6, 3)

Answer 1

Its D 6,3 YW

Explanation

Answer 2

`\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ 4p(y- k)=(x- h)^2 \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (y-3)^2=8(x-4)\implies (y-\stackrel{k}{3})^2=4(\stackrel{p}{2})(x-\stackrel{h}{4})~~ \begin{cases} \stackrel{vertex}{(4,3)}\\ p=2 \end{cases}`

now, is worthy noticing that the squared variable is the "y", and thus this is a horizontal parabola.

so let's see, the (h,k) pair for the vertex are at (4,3), and we have a positive "p" of 2 units, now, when "p" is positive, it means the parabola is opening to the right-hand-side.

So if we move horizontally from (4,3) over to 2 units to the right, we'll be landing at the focus point of (6, 3).