SOLUTION: A laboratory designed a radio telescope with a diameter of 320 feet and a maximum depth of 50 feet. The graph shows a parabola basing threw points (-160,50)(0,0)(160,50). What woul

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A laboratory designed a radio telescope with a diameter of 320 feet and a maximum depth of 50 feet. The graph shows a parabola basing threw points (-160,50)(0,0)(160,50). What woul      Log On


   

Question 883736: A laboratory designed a radio telescope with a diameter of 320 feet and a maximum depth of 50 feet. The graph shows a parabola basing threw points (-160,50)(0,0)(160,50). What would be the equation for the parabola
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Since it goes through the origin, the parabola has the form, y=ax%5E2
50=a%28160%29%5E2
a=50%2F160%5E2=50%2F25600=1%2F512
highlight%28y=x%5E2%2F512%29