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Math Help - need help with 10 questions

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ladventchildp View Post
    question number 8 doesnt really make sense isnt it suppose to be polar coordinates? what i tried to do for xy=1 is uhmm x and y has to rather equal pi/4 or 5pi/4(cuz its int he 3rd quardnant) from there i did tan(@)=1 @=pi/4 or 5pi/4 x=rcos x=1 therefore 1=rcos(@) 1=r(1/sqrt2) r=+-sqrt2 and for sin i got pi/4+npi
    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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  2. #17
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    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
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ladventchildp View Post
    and


    8)
    Change the equation to polar coordinates.
    x2 + y2 8y = 0
    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
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  4. #19
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    Quote Originally Posted by ladventchildp View Post
    k can you guys check if im doing something wrong
    for question 1
    x=6y^2
    y^2=x/6
    y2=4px
    4px=1/6 therefore p=1/24
    therefore focus h+p=(1/24,0)
    axis= 0 since k=0
    directrix=h-p=-1/24

    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)

    question 3
    x2/25-y2/4=1
    y2=4+(4x2/25)
    y=+-(2sqrt(x2+25))/5

    and i dont know if that polar coordinate link is applicable with 8 im still working on it
    can check if these answers are right before i submit it?
    thatll be great thanks
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    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...

    Quote Originally Posted by ladventchildp View Post
    k can you guys check if im doing something wrong
    for question 1
    x=6y^2
    y^2=x/6
    y2=4px
    4px=1/6 therefore p=1/24
    therefore focus h+p=(1/24,0)
    axis= 0 since k=0
    directrix=h-p=-1/24
    this looks fine


    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)
    for the foci i got (0, +/- sqrt(84)/5) ........check my calculations. for (x^2)/(b^2) + (y^2)/(a^2) = 1 where a >= b > 0 the foci is (0, +/- c) where c^2 = a^2 - b^2


    question 3
    x2/25-y2/4=1
    y2=4+(4x2/25)
    y=+-(2sqrt(x2+25))/5
    for asymptote:

    y = +/- (b/a)x = +/- (2/5)x
    Last edited by Jhevon; May 6th 2007 at 02:27 PM.
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