Conic Sections: Application Problems (5)

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Question

A parabolic dish antenna has a depth of 2020 cm and a diameter of 8080 cm at the rim. Where should the receiver be placed so that all incoming parallel rays converge there?

Solution — Step by Step

Place the vertex at origin and the axis of symmetry along the positive yy-axis. The parabola opens upward: x2=4pyx^2 = 4py, where pp is the focal distance (vertex to focus).

The rim is at depth 2020 cm and diameter 8080 cm. So at y=20y = 20, x=40x = 40 (radius =40= 40 cm).

402=4p201600=80pp=20cm40^2 = 4p \cdot 20 \Rightarrow 1600 = 80p \Rightarrow p = 20 \, \text{cm}

The focus is at (0,p)=(0,20)(0, p) = (0, 20) cm — exactly at the depth of the rim, on the axis of symmetry.

Place the receiver at the focus, 2020 cm from the vertex along the axis. Rays parallel to the axis reflect through the focus by the optical property of parabolas.

Final answer: receiver at the focus, 2020 cm from the vertex along the axis of the dish.

Why This Works

Every parabola has a reflective property — rays parallel to the axis reflect through the focus, and conversely, rays from the focus reflect parallel to the axis. This is why parabolic dishes are used for satellite TV (focus collects signal) and headlights (focus emits light, reflects parallel).

The equation x2=4pyx^2 = 4py encodes this: pp is the distance from vertex to focus.

Alternative Method

Use the directrix-focus property. For any point (x,y)(x, y) on the parabola, distance from focus = distance from directrix. Plug in the rim point (40,20)(40, 20) and solve for pp. Same answer.

For real-world parabola problems: identify “depth” with yy value at rim, “radius” with xx value at rim, and use x2=4pyx^2 = 4py. The focus is always at (0,p)(0, p) when the vertex is at origin.

Students sometimes use the diameter as xx in the equation instead of the radius. Diameter is 8080 cm; only half of it (4040 cm) is the xx-coordinate of one rim point.

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