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Question

A parabolic heater has a cross section modeled by a parabola with vertex (0,0) and focus (4,0). The heater has a depth of 8 inches. What is the width of the parabolic heater at its widest point? Round your answer to the nearest hundredth.

Answer 1

If you draw this out you will see this is a parabola lying on its side opening out to the right. The conic equation therefore is 4p(x-h) = (y-k)^2 where h and k are the coordinates of the vertex and p is the focal distance (4) which gives us (4)(4)(x)= y^2 y^2 = 16x

You will also see from inspection that as x increases so does y. You can graph this in desmos . So the widest point will be at x= 8 y^2 = 16 x 8 y^2 =128 y= 13.3137

since the parabola is symetric the total width will be twice this or 22.63 rounded to the nearest hundreth

Answer 2

Equation of parabola y² = 4px = 4·4x; y² = 16x. If x = 8, then y² = 16·8; y = ± √128 = ± 8√2.

So width is 2(8√2) = 16√2 cm = 22.63 cm