yeah, the question said polar coordinates, but i believe it should have said "polar equation" since we were not given coordinates, we were given an equation. therefore, i just replaced x with rcos(theta) and y with rsin(theta) and simplified

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- May 6th 2007, 11:41 AMJhevon
- May 6th 2007, 01:27 PMladventchildp
that makes sense, hey can you help me with second part of number 8 that one isnt working for me i dont know why but i need the coordinates and can you check my answers for 1-4 that i posted earlier thatll be great

thanks - May 6th 2007, 01:33 PMJhevon
again, just replace x with rcos(theta) and y with rsin(theta) and i'm tired of typing theta, i'll use t from now on

x^2 + y^2 - 8y = 0

=> (rcos(t))^2 + (rsin(t))^2 - 8(rsin(t)) = 0

=> r^2*cos^2(t) + r^2*sin^2(t) - 8rsin(t) = 0

=> r^2*(cos^2(t) + sin^2(t)) - 8rsin(t) = 0

=> r^2 - 8rsin(t) = 0

and i'd leave it there - May 6th 2007, 02:17 PMladventchildp
- May 6th 2007, 02:53 PMJhevon
i find questions like these so annoying i don't know why. whenever i see "find the directrix and focus of a parabola", i cringe. anyway...

this looks fine

Quote:

question 2

25x2+4y2-16=0

25x2/4^2+y2/2^2=1

a2+b1=c2 16+4=c2 2sqrt5=c therefore foci (+-2sqrt5,0) verticies(0,+-2)

Quote:

question 3

x2/25-y2/4=1

y2=4+(4x2/25)

y=+-(2sqrt(x2+25))/5

y = +/- (b/a)x = +/- (2/5)x