Page 1 of 2 12 LastLast
Results 1 to 15 of 20

Math Help - need help with 10 questions

  1. #1
    Newbie
    Joined
    Apr 2007
    Posts
    14

    need help with 10 questions

    1)
    Give the focus, directrix, and axis of the parabola.
    x = 6y2

    2)
    Find the foci and vertices of the ellipse.
    25
    x2 + 4y2 16 = 0

    3)
    Find the asymptotes of the hyperbola.
    x^2/25-y^2/4=1

    4)
    Find
    θ so that rotating axes the through an angle of θ produces a new equation containing no xy-term.
    31x2 + 10sqrt(3xy) + 21y2 −144 = 0

    5)
    Use the angle found in problem 4
    to rotate the conic and eliminate the xy term. Write the equation of
    rotated conic.
    31x2 + 10sqrt(3xy) + 21y2 −144 = 0

    6)

    Identify the conic produced and the angle of rotation from problem 4
    31x2 + 10sqrt(3xy) + 21y2 −144 = 0

    7)
    Find the rectangular coordinates for the point.
    (
    3,2pi/3)

    8)
    Change the equation to polar coordinates.
    xy = 1

    and
    x2 + y2 8y = 0


    9)
    Change the equation to rectangular coordinates.

    r = 2
    and
    r=sin(theta)

    10)
    Find a rectangular equation for the plane curve defined by the parametric equations.
    x = 4sin θ, y = cos θ

    and
    x
    = tan(t), y = sec(t)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,719
    Thanks
    634
    Hello, ladventchildp!

    If you're doing Rotations, you must be familiar with certain formulas.


    4. Find θ so that rotating axes the through an angle of θ
    produces a new equation containing no xy-term.
    . . 31x + (10√3)xy + 21y − 144 .= .0
    The general quadratic equation is: .Ax + Bxy + Cy + Dx + Ey + F .= .0
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B
    The angle of rotation θ is given by: .tan(2θ) .= .------
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A - C

    . . . . . . . . . . . . . . . . . . . _
    We have: .A = 31, B = 10√3, C = 21, D = 0, E = 0, F = -144
    . . - . . - . . . . . . . . . . . _
    . . . . . . . . . . . . . . .10√3 . - - . _
    Hence: . tan2θ .= .--------- .= .√3
    . . . . . . . . . . . . . .31 - 21

    Then: . .= .π/3 . . . . θ .= .π/6



    5) Use the angle found in problem 4 to rotate the conic and eliminate the xy term.
    Write the equation of rotated conic.
    To eliminate the xy-term, we use these substitutions:
    . . x' .= .xcosθ - ysinθ
    . . y' .= .xsinθ + ycosθ

    With θ = π/6, we have:
    . . x .= .√3x - y
    . . y .= .x + √3y


    Substitute:

    31(√3x - y) + 10√3(√3x - y)(x + √3y) + 21(x + √3y) - 144 .= .0

    This simplifies to: .36x + 16y .= .144



    6) Identify the conic produced and the angle of rotation from problem 4
    . . . . . . . . . . . . . . . . . .x . . y
    The conic is an ellipse: . --- + --- .= .1
    . . . . . . . . . . . . . . . . . .4 . . .9

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We could have identitied the conic long ago.

    The equation has a discriminant: .d .= .B - 4AC . .
    (Look familiar?)
    . . If d = 0: parabola
    . . If d > 0: hyperbola
    . . If d < 0: ellipse

    . . . . . . . . . . . . . . . . . . . _
    We have: .A = 31, B = 10√3, C = 21
    . . - . . . - . . . - . . . . _
    Therefore: .d .= .(10√3) - 4(31)(21) .= .-2304 . . . . ellipse

    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Soroban View Post
    Hello, ladventchildp!

    If you're doing Rotations, you must be familiar with certain formulas.


    The general quadratic equation is: .Ax + Bxy + Cy + Dx + Ey + F .= .0
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B
    The angle of rotation θ is given by: .tan(2θ) .= .------
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A - C

    . . . . . . . . . . . . . . . . . . . _
    We have: .A = 31, B = 10√3, C = 21, D = 0, E = 0, F = -144
    . . - . . - . . . . . . . . . . . _
    . . . . . . . . . . . . . . .10√3 . - - . _
    Hence: . tan2θ .= .--------- .= .√3
    . . . . . . . . . . . . . .31 - 21

    Then: . .= .π/3 . . . . θ .= .π/6



    To eliminate the xy-term, we use these substitutions:
    . . x' .= .xcosθ - ysinθ
    . . y' .= .xsinθ + ycosθ

    With θ = π/6, we have:
    . . x .= .√3x - y
    . . y .= .x + √3y


    Substitute:

    31(√3x - y) + 10√3(√3x - y)(x + √3y) + 21(x + √3y) - 144 .= .0

    This simplifies to: .36x + 16y .= .144



    . . . . . . . . . . . . . . . . . .x . . y
    The conic is an ellipse: . --- + --- .= .1
    . . . . . . . . . . . . . . . . . .4 . . .9

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We could have identitied the conic long ago.

    The equation has a discriminant: .d .= .B - 4AC . .
    (Look familiar?)
    . . If d = 0: parabola
    . . If d > 0: hyperbola
    . . If d < 0: ellipse

    . . . . . . . . . . . . . . . . . . . _
    We have: .A = 31, B = 10√3, C = 21
    . . - . . . - . . . - . . . . _
    Therefore: .d .= .(10√3) - 4(31)(21) .= .-2304 . . . . ellipse

    Congrats on your 18th post!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2007
    Posts
    14
    i got up to the simplifying part just having a hard time simplifying heh
    thanks alot

    well it seems i got everything besides 7-10 if you can help me with those thatll be great im working on it, its just taking a while sigh
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ladventchildp View Post
    i got up to the simplifying part just having a hard time simplifying heh
    thanks alot

    well it seems i got everything besides 7-10 if you can help me with those thatll be great im working on it, its just taking a while sigh
    See my first post here http://www.mathhelpforum.com/math-he...ordinates.html for the rules

    I'll do one of your questions as an example, try the rest and tell me if you get it

    7) (3, 2pi/3) to rectangular coordinates

    => (r, theta) = (3, 2pi/3)

    x = rcos(theta) = 3cos(2pi/3) = -3/2
    y = rsin(theta) = 3sin(2pi/3) = 3sqrt(3)/2

    so the rectangular coordinates are: (-3/2 , 3sqrt(3)/2)

    now try the other questions of this type (8 and 9)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2007
    Posts
    14
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ladventchildp View Post


    10)
    Find a rectangular equation for the plane curve defined by the parametric equations.
    x = 4sin θ, y = cos θ

    we start by solving for the paramentric variable in terms of x, then we replace the parametric variable with the expression in y

    x = 4sin(theta)
    => sin(theta) = x/4 .............remember this, we'll get back to it
    => theta = arcsin(x/4)

    now y = cos(theta)
    => y = cos(arcsin(x/4))
    now recall that arcisn(x/4) = theta and sin(theta) = x/4
    so we draw a right triangle with one acute angle being theta, the length of the side opposite to the angle is x and the hypotenuse is length 4. by pythagoras, the adjacent side is sqrt(16 - x^2), from this triangle we see that:

    y = cos(arcsin(x/4)) = cos(theta) = sqrt(16 - x^2)/4
    => 4y = sqrt(16 - x^2)
    => 16y^2 = 16 - x^2
    => x^2 + 16y^2 = 16
    => (x^2)/16 + y^2 = 1
    this is an ellipse. now try the other part of this question
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2007
    Posts
    14
    k i used a different method to solving number 10 buti got it
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ladventchildp View Post
    k i used a different method to solving number 10 buti got it
    what method was that?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Apr 2007
    Posts
    14
    well what i did was x=tan(t) y=sec(t)
    1+tan^2=sec^2
    1+x^2=y^2
    y^2-x^2=1
    hyperbola
    similarily x=4sin(@)
    x/4=sin(@)
    sin^2+cos^2=1
    (x/4)^2+y^2=1 same answer though can you help me with
    number 8 and second part of number 9
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ladventchildp View Post
    well what i did was x=tan(t) y=sec(t)
    1+tan^2=sec^2
    1+x^2=y^2
    y^2-x^2=1
    hyperbola
    similarily x=4sin(@)
    x/4=sin(@)
    sin^2+cos^2=1
    (x/4)^2+y^2=1 same answer though can you help me with
    number 8 and second part of number 9
    that's nice! i never thought of using identities. i should remember that technique for the future
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ladventchildp View Post


    8)
    Change the equation to polar coordinates.
    xy = 1

    and
    x2 + y2 − 8y = 0

    remember, x = rcos(theta) and y = rsin(theta)
    so xy = 1
    => rcos(theta)*rsin(theta) = 1
    => r^2 * sin(theta)cos(theta) = 1
    => r^2 = 1/(sin(theta)cos(theta))
    => r^2 = sec(theta)csc(theta)



    9)
    Change the equation to rectangular coordinates.


    r=sin(theta)


    r = sin(theta)
    => r^2 = rsin(theta)
    => x^2 + y^2 = y
    => x^2 + y^2 - y = 0
    => x^2 + y^2 - y + (-1/2)^2 = (-1/2)^2
    => x^2 + (y - 1/2)^2 = (1/2)^2
    this is a circle with center (0, 1/2) and radius 1/2
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Apr 2007
    Posts
    14
    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
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Apr 2007
    Posts
    14
    and for 9 how did you get one half can you walk me through that problem again, it sort of doesnt make sense
    Follow Math Help Forum on Facebook and Google+

  15. #15
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ladventchildp View Post
    and for 9 how did you get one half can you walk me through that problem again, it sort of doesnt make sense
    all i did was complete the square for y, do you see it now? i had to add the square of half the coefficient of y, and because we're dealing with an equation, i had to add it to the other side as well, to keep things equal
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 04:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 04:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 03:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 07:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum