Matteo is adjusting a satellite because he finds it is not focusing the incoming radio waves perfectly. The shape of the satellite can be modeled by [tex]y^2=6 x[/tex] where [tex]x[/tex] and [tex]y[/tex] are modeled in inches. He 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. [tex](-1.5,0)[/tex]
B. [tex](0,-1.5)[/tex]
C. [tex](0,1.5)[/tex]
D. [tex](1.5,0)[/tex]
Answer (1)
The problem involves finding the focus of a parabola given its equation. The equation is in the form y 2 = 6 x .
Compare the given equation with the standard form y 2 = 4 a x .
Determine that 4 a = 6 .
Solve for a to find a = 1.5 .
State the focus point as ( 1.5 , 0 ) .
Explanation
Problem Analysis The equation of the satellite dish is given by y 2 = 6 x . This is a parabola opening to the right. We need to find the focus of this parabola.
General Form of a Parabola The general form of a parabola opening to the right is y 2 = 4 a x , where a is the distance from the vertex to the focus. The vertex is at the origin ( 0 , 0 ) , and the focus is at ( a , 0 ) .
Comparing Equations Comparing the given equation y 2 = 6 x with the general form y 2 = 4 a x , we have 4 a = 6 .
Solving for a To find the value of a , we solve the equation 4 a = 6 for a . Dividing both sides by 4, we get a = 4 6 = 2 3 = 1.5 .
Finding the Focus Therefore, the focus of the parabola is at ( 1.5 , 0 ) .
Examples
Understanding the focus of a parabola is crucial in designing satellite dishes and antennas. The focus point is where incoming parallel rays converge after reflecting off the parabolic surface. In a satellite dish, the receiver is placed at the focus to maximize signal reception. Similarly, in a flashlight, the light source is placed at the focus of a parabolic reflector to produce a parallel beam of light. This principle ensures efficient transmission or reception of signals and energy.
