Please answer this analytical geometry problem?

Question

Problem 3: Locating Whales
Hyperbolas can be used to locate objects underwater. To locate a whale in the ocean, two microphones are placed 8000 feet apart. One microphone picks up a whale noise 0.4 seconds after the second microphone picks up the same noise. The speed of sound in water is about 5000 feet per second.
a. How much farther from the whale is the first microphone?
b. Find an equation for the possible locations of the whale.
c. What is the closest distance that the whale could be to the second microphone?
d. Will the whale always be closer to the microphone that receives the signal first? Can the whale be on either branch of the hyperbola? Explain your reasoning.

Answer 1

The sound waves bounce of the whale and come back. Now the waves have travelled for 0.4 seconds more or 2000 feet more. so the further microphone is 1000 feet further from the whale.

Answer 2

concentration (4, 0): Fx = 4 Fy = 0 Directrix: x = 0 because the directrix is a function of x, the equation will be x as a function of y, so p = a million/2(Fx - x) p = a million/2(4 - 0) p = a million/2(4) p = 2 Vertex (Fx - p, Fy) Vertex (4 - 2, 0) Vertex (2, 0) h = 2 ok = 0 a = a million / 4p a = a million / 4(2) a = a million/8 x = a(y - ok)² + h x = a million/8(y - 0)² + 2 x = a million/8 y² + 2 ¯¯¯¯¯¯¯¯¯¯¯¯ concentration (2, 4): Fx = 2 Fy = 4 Directrix: y = - 2 because the directrix is a function of y, the equation will be y as a function of x, so p = a million/2(Fy - y) p = a million/2(4 - (- 2)] p = a million/2(4 + 2) p = a million/2(6) p = 3 Vertex (Fx, Fy - p) Vertex (2, 4 - 3) Vertex (2, a million) h = 2 ok = a million a = a million / 4p a = a million / 4(3) a = a million/12 y = a(x - h)² + ok y = a million/12(x - 2)² + a million y = a million/12(x² - 4x + 4) + a million y = a million/12 x² - a million/3 x + a million/3 + 3/3 y = a million/12 x² - a million/3 x + 4/3 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Answer 3

After waiting for an hour, you probably answered this yourself ... but here goes for part of it

a) the difference in distance is 0.4 x 5000 feet = 2000 feet
c) 3000 and 5000 = 8000 with a diff of 2000 (that is if it in a straight line
d) using maths, yes ... using physics, no
the slower signal may need to rebound to get to the microphone or the tide may favour the further mike, making it seem closer