In the given figure pt is the bisector of angle qpr. qs=3x, rs=2x+2, pq=3y-1 and pr=y+5
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Refer to the diagram shown below. It shows a vertical cross-section of the paraboloid through its axis of symmetry.
Let the vertex of the parabola be at the origin. Then its equation is of the form
y = bx²
Because the parabola passes through (18,8), therefore
8 = b(18²)
b = 0.02469
The parabola is y = 0.02469x².
The receiver should be placed at the focal point of the paraboloid for optimal reception.
The y-coordinate of the focus is
a = 1/(4b) = 1/0.098765 = 10.125 in
Answer: The receiver is located at 10.125 inches from the vertex.
Let the vertex of the parabola be at the origin. Then its equation is of the form
y = bx²
Because the parabola passes through (18,8), therefore
8 = b(18²)
b = 0.02469
The parabola is y = 0.02469x².
The receiver should be placed at the focal point of the paraboloid for optimal reception.
The y-coordinate of the focus is
a = 1/(4b) = 1/0.098765 = 10.125 in
Answer: The receiver is located at 10.125 inches from the vertex.