A parabolic heater has a cross section modeled by a parabola with vertex (0,0\right) and focus (2,0\right) . The heater has a depth of 5 inches. What is the width of the parabolic heater at its widest point? Round your answer to the nearest hundredth.

y = 6.93

Step-by-step explanation:

Given the Vertex and ficuas of a parabola as :

Vertex (0,0) ; focus(0,2) ; depth = 5 inches

Width of parabolic heater at its widest point :

General equation of a parabola:

4p(x - h) = (y - k)²

(h, k) = Vertex

hence,

h =0 ; k = 0 ; p = 2 ; x = 5

(y - k)² = 4p(x - h)

(y - 0)² = 4*2(6 - 0)

(y-0)(y-0) = 8(6 - 0)

y² = 48 - 0

y² = 48

y = 6.928

y = 6.93

.

step-by-step explanation:

.

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation