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A radio telescope has a parabolic surface, as shown below. A parabola opening up with vertex at the origin is graphed on the coordinate plane. The height of the parabola from top to bottom is 9 meters and its width from left to right is 12 meters. If the telescope is 9 m deep and 12 m wide, how far is the focus from the vertex?

Sagot :

The telescope shape and the characteristic equations of the telescope parameters are the same as parabolic equations

The distance between the focus and the vertex, of the parabola is 3.375 meters

The process for obtain the above values is as follows:

The known parameters of the parabola are;

The location of the vertex of the parabola= The origin = (0, 0)

The height of the parabola = 9 meters

The width of the parabola = 12 meters

The unknown parameter;

The distance between the focus and the vertex

Method:

Finding the coordinate of the focus from the general equations of the the parameters of a parabola

The equation of the parabola in standard form is y = a·(x - h)² + k

From which we have;

(x - h)² = 4·p·(y - k)

The coordinates of the focus, f = (h, k + p)

Where;

(h, k) = The coordinates of the vertex of the parabola = (0, 0)

a = 1/(4·p)

From the question, we have the following two points on the parabola,

given that the parabola is 12 meters wide at 9 meters above the origin and

it is symmetric about the y-axis;

Points on the parabola = (9, 6), and (9, -6)

Plugging in the values of the vertex, (h, k) and the two known points, in the equation, y = a·(x - h)² + k, we get;

6 = a·(9 - 0)² + 0 = 81·a

a = 6/81 = 2/27

p = 1/(4·a)

∴ p = 1/(4 × 2/27) = 27/8

The coordinate of the focus, f =  (h, k + p)

∴ f = (0, 0 + 27/8) = (0, 27/8)

The coordinate of the focus f = (0, 27/8)

Given the vertex and the focus of the parabola have the same x-values of 0, we have;

The distance between the focus and the vertex, d = the difference in their y-values;

d = 27/8 - 0 = 27/8 = 3.375

The distance between the focus and the vertex, d = 3.375 meters

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