A satellite dish is the shape of a paraboloid. The dish is 30 inches wide, and 10 inches deep. How many inches should
the receiver be located from the vertex for optimal reception?
(round to the nearest thousandth)
Receiver •

The distance that the receiver should be located from the vertex to receive optimal reception is;

5.625 inches above the vertex

We are given;

Width of the satellite dish = 30 inches

Depth of the satellite dish = 10 inches

Vertex of parabola is at (0, 0)

Thus, we can say that this parabola passes through the point (15, 10).

Now, the conic form of a parabola with its vertex at the origin is given by the formula; y = (1/4c)x²

where c is the distance between the focus and the vertex.

Plugging in the relevant values gives;

10 =  (1/4c)(15)²

4c = 225/10

4c = 22.5

c = 22.5/4

c = 5.625

Thus, the receiver should be located at 5.625 inches above the vertex so for optimal reception.

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The receiver should be located at (0, 5.625)

Step-by-step explanation:

The given parameters are;

The width of the satellite dish = 30 inches

The depth of the dish = 10 inches

The receiver should be located at the focus of the parabola

Where the vertex of a parabola is (0, 0)

The equation of an upward or downward facing parabola is x² = 4·p·y

The focus of the parabola = (0, p)

The directrix is given by y = -p

The axis is the x-axis

Therefore, given that the width = 30 inches and the depth = 30 inches, we have;

The points (-15, 10), (15, 10), and (0, 0) (which is the vertex) on the parabola

Substituting, the coordinates into the equation of the parabola gives;

(-15)² = 4 × p × 10

p = (-15)²/(4 × 10) = 5.625

The coordinate of the focus = (0, 5.625)

Therefore, the receiver should be located at (0, 5.625).

4sqrt5

step-by-step explanation:

the base has dimensions of 8 x 4, so the diagonal is 64 + 16 square rooted or 4sqrt5.