1. ## Parabola word problem

A satellite is placed in a position to make a parabolic flight past the moon with the center of the moon at the focal point of the parabola. When the satellite is 5783 km from the surface of the moon, it makes an angle of 60 degrees with the axis of the parabola. The vertex of the parabola is 143 km from the surface of the moon. Use the information given to compute the diameter of the moon.

Thank you

2. Originally Posted by Rae
A satellite is placed in a position to make a parabolic flight past the moon with the center of the moon at the focal point of the parabola. When the satellite is 5783 km from the surface of the moon, it makes an angle of 60 degrees with the axis of the parabola. The vertex of the parabola is 143 km from the surface of the moon. Use the information given to compute the diameter of the moon.

Thank you
1. Define a coordinate system such that the vertex of the parabola is at V(0, 0) and the parabola is opening up.

2. Draw a sketch. (see attachment)

3. Let r denote the radius of the moon. Then the equation of the parabola in question is $\displaystyle x^2=4py$ with the focus at F(0, 143+r) . That means $\displaystyle p = 143+r$

4. Since $\displaystyle \cos(60^\circ)=\frac12$ and $\displaystyle \sin(60^\circ)=\frac12\cdot \sqrt{3}$ the satellite has the coordinates $\displaystyle S\left((5783+r)\cdot \frac12 \cdot \sqrt{3}\ ,\ 143+r+(5783+r) \cdot \frac12 \right)$

5. Plug in the value of p and the coordinates of S into the equation of the parabola:

$\displaystyle \left((5783+r)\cdot \frac12 \cdot \sqrt{3} \right)^2=4\cdot (r+143) \left(143+r+(5783+r) \cdot \frac12\right)$

6. After moving some stuff around (a lot of stuff btw) you'll get a quadratic equation in r:

$\displaystyle 7r^2+5762r-31128777=0$

which yields 2 solutions. The negative solution isn't very plausible here. so it is

$\displaystyle r = 1737\ km$

3. Great, thanks a lot!