Venus orbits the sun in an approximately circular path. Venus is about 67 million miles from the sun. A comet is located 80 million miles north and 73 million miles west of the sun. The comet follows a straight-line path and exits Venus's orbit at the east most edge.

Question

a. Find the location where the comet enters Venus's orbit to the nearest tenth?

b. What is the closest distance the comet comes to the sun.

c. If the comet travels at a constant speed of 0.02 million miles per hour, then how long does the comet stay in Venus's orbit

I don't understand any of this to be honest, we have to use secant, perpendicular lines, and/or points of intersections.

Answer 1

let's put the sun at the origin and use x and y in units of 1 million
so the orbit of Venus can be written appr a
x^2 + y^2 = 67^2
then the comet is at (-73,80) which is clearly outside of the orbit of Venus.

the most eastern point on Venus' orbit is (67,0)

it might help to make a sketch here

A line cutting a circle at two points is called a secant
slope of our line = (0-80)/(67+73) = -80/140 = -4/7
equation of line: y = (-4/7)x + b
with (67,0) on it
0 = (-4/7)(67) + b
b = 268/7 ----> y = (-4x + 268)/7

sub that into the equation of the circle:
x^2 + ( (-4x + 268)/7)^2 = 4489
x^2 + (16x^2 - 2144x + 71824)/49 = 4489
times 49
49x^2 + 16x^2 - 2144x + 71824 = 219961
65x^2 - 2144x - 148137 = 0
arrgggh!!!
but it is not so bad, since we know x = 67 works, so
(65x + 2211)(x - 67) = 0
x = -2211/65 or x = 67

a) x = -2211/65 or appr -34.0154, used my calculator to find y = appr 57.7231

(-34.0154 , 57.7231) convert that to suit the question, (multiply by 1,000,000)

verify:
http://www.wolframalpha.com/input/?i=y+%3D+%28-4x+%2B+268%29%2F7+%2C+x%5E2+%2B+y%5E2+%3D+67%5E2

b) closest of (0,0) to y = (-4x + 268)/7
7y = -4x + 268
4x + 7y - 268 = 0
distance to (0,0) = |0+0-268|/√(16+49) = appr 33.2413 million

c) find length of line from (-34.0154,57.7231) to (67,0)
since time = distance/rate .....

carry on

Answer 2

Can you see how this question is the same as
the one that Steve answered for you here ?

http://www.jiskha.com/display.cgi?id=1443993008

Only the numbers have been changed to protect the innocent.

Answer 3

Oh wow, thank you so much I actually got that, just I didn't know how to continue on from there.. but yay now I know how to!!! Thank you so much, I really thought that equation was impossible. Thanks for making it clear, once again(: