Answer:
The antenna should be place 0.75 inches above the vertex.
Step-by-step explanation:
We have the equation of the parabola:
(x-4)²=3(y-3)
The general equation of the parabola is:
(x-x₀)²=4p(y-y₀)
where:
(x₀,y₀) is the vertex of the parabola
p is the focus of the parabola
Hence,
The vertex of the parabola is (4,3)
The focus of the parabola is given by:
[tex]3=4p\\ p=\frac{3}{4}\\ p=0.75[/tex]
Then, we have to shift the antenna towards the focus point. The focus point would be:
(x₀,y₀+p)=(4,3+0.75)= (4,3.75)
The directrix of the parabola is:
y=y₀-p
y=3-.75=2.25
A graph of the parabola is attached, with its focus point (4.3.75) and its directrix (y = 2.25)
Answer:
(4,3.75).
Step-by-step explanation:
The shape of his satellite can be modeled by the given equation is
[tex](x-4)^2=3(y-3)[/tex] .... (1)
where x and y are modeled in inches.
He realizes that the static is a result of the feed antenna shifting slightly off the focus point. So, we need to find the focus of the given parabola.
If the equation of a parabola is
[tex](x-h)^2=4p(y-k)[/tex] .... (2)
then focus of the parabola is (h,k+p).
On comparing (1) and (2) we get
[tex]h=4,k=3, 4p=3\Rightarrow p=\dfrac{3}{4}[/tex]
[tex]Focus=(h,k+p)=(4,3+\dfrac{3}{4})=(4,3+0.75)=(4,3.75)[/tex]
Therefore, the feed antenna should be placed at (4,3.75).
