Results 1 to 7 of 7

Math Help - Curve's point A tangent line passes through a point P. What is the distance of AP

  1. #1
    Junior Member
    Joined
    May 2011
    Posts
    40

    Curve's point A tangent line passes through a point P. What is the distance of AP

    Let A be a point on the curve C: {x}^{ 2} + {y}^{ 2}  - 2x + 4 = 0. If the tangent line at A passes through P(4,3), then the length of AP is?

    So i'm stuck midway at the solution for this one.
    So far I've gotten these equations
    {y}^{ '} = \frac{1-x}{ y} = m ; the tangent line slope at any point on the curve (by implicit differentiation)
    m = \frac{y-3}{x-4} ; the slope at point P; acquired by y-k=m(x-h)

    and when I put those two equations together I get:
    {-x}^{ 2} + 5x - 4 + 3y= {y}^{ 2}

    And when I input this equation back into the original curve i end up with
    x= -y

    I'm stuck at this point because when I plug in the last equation back to original curve again, I'm getting an complex solution.
    Help please!

    Oh Gosh. I am sorry. That's supposed to be
    {x}^{ 2} + {y}^{ 2}  - 2x - 4 = 0
    Last edited by gundanium; May 15th 2011 at 10:19 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    You sure you got the curve C right. Completing the square in x gives

    (x-1)^2 + y^2 + 3 = 0

    There are no real solutions to this equation.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2011
    Posts
    40
    Quote Originally Posted by Danny View Post
    You sure you got the curve C right. Completing the square in x gives

    (x-1)^2 + y^2 + 3 = 0

    There are no real solutions to this equation.
    Oh Gosh. I am sorry. That's supposed to be
    {x}^{ 2} + {y}^{ 2} - 2x - 4 = 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by gundanium View Post
    Let A be a point on the curve C: {x}^{ 2} + {y}^{ 2}  - 2x - 4 = 0. If the tangent line at A passes through P(4,3), then the length of AP is?

    ...
    1. Your curve is a circle with the center M(1, 0) and the radius r = \sqrt{5} .

    2. The triangle MPA is a right triangle with MP as the hypotenuse and MA = r.

    3. Use Pythagorean theorem to determine the length of AP.
    Spoiler:
    (MP)^2=(AP)^2+r^2~\implies~|\overline{AP}|=\sqrt{1  3}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    This is the way I would do it. Suppose the point on the circle is (a,b).  Then (a,b) satisfies

    a^2+b^2-2a-4 = 0\;\;\;(1).

    The tangent to the curve at this point is

    m = \dfrac{1-a}{b}

    and the slope of the line connecting (a,b) to (3,4) is

    m = \dfrac{b-3}{a-4}

    These must equal, hence

     \dfrac{b-3}{a-4} =  \dfrac{1-a}{b}

    or

    a^2+b^2 - 5a - 3b = -4\;\;\;(2)

    You are required to find the distance between the two points or

    d = \sqrt{(a-4)^2+(b-3)^2}
    = \sqrt{a^2+b^2 - 8a -6b +25}

    Now 2\times(2)-(1) gives

    a^2+b^2 - 8a - 6b + 12 = 0

    so

    d = \sqrt{25-12} = \sqrt{13}

    Edit: Earthboth's solution is so much more elegant (and of course easier!)
    Last edited by Jester; May 16th 2011 at 05:47 AM. Reason: fixed typo
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2011
    Posts
    40
    Quote Originally Posted by Danny View Post
    Now 2\times(2)-(1) gives
    Why? How did you come up with this?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    This is what I have

    a^2+b^2-2a = 4\;\;\;(1)

    a^2+b^2 - 5a - 3b = -4\;\;\;(2)

    this is what I want

     a^2+b^2 -8a - 6b\;\;\;(3)

    Multiply p \times (1) + q \times (2) = (3) (LHS only) and equate like terms. As it turns out, there is a unique p and q solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: February 9th 2010, 06:43 PM
  2. Replies: 3
    Last Post: December 12th 2009, 02:53 PM
  3. Replies: 5
    Last Post: November 4th 2009, 07:49 PM
  4. Replies: 2
    Last Post: August 18th 2009, 06:55 AM
  5. Replies: 13
    Last Post: August 3rd 2009, 08:28 PM

Search Tags


/mathhelpforum @mathhelpforum