Results 1 to 6 of 6

Math Help - tangent with "unknown parabola"

  1. #1
    Junior Member
    Joined
    Apr 2010
    From
    Australia
    Posts
    26

    tangent with "unknown parabola"

    im sorry for posting so many questions. this is the last.

    "the line  y=5x-3 is a tangent to the parabola  y=ax^2 + bx -1 at the point (1,2). find a and b"

    now i have no idea where to start here. i could assume solving simultaneuosly could give some sort of answer but it would contain both unknowns.

    if i differentiate y=ax^2 + bx -1 i get
    y'= 2ax+b
    and from the tangent we know that
    2ax + b = 5 at (1,2)

    its here that im confused. whether the whole previous section is wrong i just cant get an answer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by link0099 View Post
    im sorry for posting so many questions. this is the last.

    "the line  y=5x-3 is a tangent to the parabola  y=ax^2 + bx -1 at the point (1,2). find a and b"

    now i have no idea where to start here. i could assume solving simultaneuosly could give some sort of answer but it would contain both unknowns.

    if i differentiate y=ax^2 + bx -1 i get
    y'= 2ax+b
    and from the tangent we know that
    2ax + b = 5 at (1,2)

    its here that im confused. whether the whole previous section is wrong i just cant get an answer.
    Just consider again what you are given, and use what you already have.
    First, because the parabola must pass through the point (1,2) you must have that
    2=a\cdot 1^2+b\cdot 1-1
    Second, since the given line with slope 5 is to be tangent to the parabola at point (1,2) you must also have that
    5=2a\cdot 1+b
    In other words, you have a system of two linear equations for a and b. - Solve it. - Done!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2010
    From
    Australia
    Posts
    26
     a\cdot 1^2+b\cdot 1-1 =2  (1)
     2a\cdot 1+b =5     (2)
    so simultaneously solved we get (2) - (1)
    a = 3
    and sub a=3 into (1)
    3 + b -1 = 2
    b-1 = -1
    b=0
    is that right

    therefore parabola eqn =
    y= 3x^2 -1

    maybe?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by link0099 View Post
     a\cdot 1^2+b\cdot 1-1 =2  (1)
     2a\cdot 1+b =5     (2)
    so simultaneously solved we get (2) - (1)
    a = 3
    No, I think you get a=2
    and sub a=3 into (1)
    3 + b -1 = 2
    b-1 = -1
    b=0
    is that right
    Not quite, you get b=1

    therefore parabola eqn =
    y= 3x^2 -1
    maybe?
    y=2x^2+x-1 would be more likely correct, I think.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2010
    From
    Australia
    Posts
    26
    oh yes 2 makes more sense. thanks a lot
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Check the calcualtions.

    a+b = 3...(1)

    2a + b = 5 ...(2)

    Now solve for a and b.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 17th 2011, 02:50 PM
  2. Replies: 2
    Last Post: April 24th 2011, 07:01 AM
  3. Replies: 1
    Last Post: October 25th 2010, 04:45 AM
  4. Solving for unknown "x" involving trig function
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: August 27th 2010, 04:59 AM
  5. Replies: 1
    Last Post: June 4th 2010, 10:26 PM

Search Tags


/mathhelpforum @mathhelpforum