- #1
rocomath
- 1,755
- 1
HINT! Parabola
An arch is in the shape of a parabola with a vertical axis. The arch is 15 ft high at the center and the 40 ft wide at the base. At which height above the base is the width 20 ft?
[tex]V(0,15)[/tex] so my equation becomes: [tex](x-0)^2=-4p(y-15)\rightarrow x^2=-4p(y-15)[/tex]
[tex]F(0,15-p)[/tex] & [tex]D:y=15+p[/tex]
I can substitute 20 into my equation, but I don't know how to find p.
Just a hint please :)
An arch is in the shape of a parabola with a vertical axis. The arch is 15 ft high at the center and the 40 ft wide at the base. At which height above the base is the width 20 ft?
[tex]V(0,15)[/tex] so my equation becomes: [tex](x-0)^2=-4p(y-15)\rightarrow x^2=-4p(y-15)[/tex]
[tex]F(0,15-p)[/tex] & [tex]D:y=15+p[/tex]
I can substitute 20 into my equation, but I don't know how to find p.
Just a hint please :)