Radio telescope parabola question

[wat2000]

If a particular radio telescope is 100ft in diameter and has a cross section modeled by the equation x^2=167y, how deep is the parabolic dish? What is the location of the focus?

can someone show me some steps to solving this? I have (167/4,0) as the focus for the second part but I am not sure if that's right. i just plugged in 4p=167 and divided to get the 167/4.

 

[lanedance]

well to start what is the defintion of a focus relative to a parabola?

for the depth you will know that 2y = 100ft

 

[wat2000]

so y = 50 ft? where do i go from there?

 

[wat2000]

could i do (1/67)x^2 with x=50 and square x and get 2500/167?

 

[ideasrule]

can someone show me some steps to solving this? I have (167/4,0) as the focus for the second part but I am not sure if that's right. i just plugged in 4p=167 and divided to get the 167/4.

Yes, that's right. The general equation for a vertical parabola is (x-h)^2=4p(y-k). Plugging in the point (1,1/167) gives you your answer.

As for the depth of the parabolic dish, draw a graph of x^2=167y. What is x when the diameter reaches 100 ft? What must y be?

 

[wat2000]

its 50 right? If so does that make the depth 50?

 

[lanedance]

if y = 50ft

then
x^2 = 167*y = 167(50)

now solve for x

 

[wat2000]

wouldnt solving for x give me x= 2500/167?

 

[ideasrule]

Yes.

 

[wat2000]

so would that give me my depth?so the focus would be 64/7 and the depth would be 14.97. and if so, what is the 8350 i got from multiplying 167 and 50?

 

[wat2000]

I get 14.97 for y and 8350 for x. do i subtract 1497 from 8350 to get the depth?