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

  1. #1
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    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: $\displaystyle {x}^{ 2} + {y}^{ 2} - 2x + 4 = 0$. 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
    $\displaystyle {y}^{ '} = \frac{1-x}{ y} = m $ ; the tangent line slope at any point on the curve (by implicit differentiation)
    $\displaystyle 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:
    $\displaystyle {-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
    $\displaystyle {x}^{ 2} + {y}^{ 2} - 2x - 4 = 0$
    Last edited by gundanium; May 15th 2011 at 09:19 AM.
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  2. #2
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    You sure you got the curve C right. Completing the square in $\displaystyle x$ gives

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

    There are no real solutions to this equation.
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  3. #3
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    Quote Originally Posted by Danny View Post
    You sure you got the curve C right. Completing the square in $\displaystyle x$ gives

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

    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
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  4. #4
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    Quote Originally Posted by gundanium View Post
    Let A be a point on the curve C: $\displaystyle {x}^{ 2} + {y}^{ 2} - 2x - 4 = 0$. 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 $\displaystyle 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:
    $\displaystyle (MP)^2=(AP)^2+r^2~\implies~|\overline{AP}|=\sqrt{1 3}$
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  5. #5
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    This is the way I would do it. Suppose the point on the circle is $\displaystyle (a,b). $ Then $\displaystyle (a,b)$ satisfies

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

    The tangent to the curve at this point is

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

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

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

    These must equal, hence

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

    or

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

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

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

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

    $\displaystyle a^2+b^2 - 8a - 6b + 12 = 0$

    so

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

    Edit: Earthboth's solution is so much more elegant (and of course easier!)
    Last edited by Jester; May 16th 2011 at 04:47 AM. Reason: fixed typo
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  6. #6
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    Quote Originally Posted by Danny View Post
    Now $\displaystyle 2\times(2)-(1)$ gives
    Why? How did you come up with this?
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  7. #7
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    This is what I have

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

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

    this is what I want

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

    Multiply $\displaystyle p \times (1) + q \times (2) = (3)$ (LHS only) and equate like terms. As it turns out, there is a unique $\displaystyle p$ and $\displaystyle q$ solution.
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