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Thread: Tangent Line Problem--Please Help

  1. #1
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    Smile Tangent Line Problem--Please Help

    Here is how the problem reads, the graphing part I got, it's just the problem itself.

    Graph the two parabolas y=x^2 and y=-x^2 +2x -5 in the same coordinate plane. Find the equations of the two lines simultaneously tangent to both parabolas.

    I got as far as just finding the derivatives which as 2x and -2x+2, but I can't the equations.
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  2. #2
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    Quote Originally Posted by doctorgk View Post
    Here is how the problem reads, the graphing part I got, it's just the problem itself.

    Graph the two parabolas y=x^2 and y=-x^2 +2x -5 in the same coordinate plane. Find the equations of the two lines simultaneously tangent to both parabolas.

    I got as far as just finding the derivatives which as 2x and -2x+2, but I can't the equations.
    Given an parabola $\displaystyle y=ax^2+bx+c,a\not = 0$ and a line $\displaystyle y=px+q$. This line is tangent to the parabola if and only if $\displaystyle (b-p)^2 - 4a(c-q)=0$. So if $\displaystyle px+q$ is tangent to $\displaystyle x^2$ then it means $\displaystyle p^2 + 4q = 0$ and it tangent to $\displaystyle -x^2+2x-5$ that means $\displaystyle (p-2)^2 - 4(q+5)=0$. The solutions to this system of equations is $\displaystyle (p,q) = \{(-2,-1),(4,-4)\}$.
    Last edited by ThePerfectHacker; Nov 11th 2007 at 08:00 AM.
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  3. #3
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    Okay but still a bit lost

    how did you get that formula? is there a way to solve this with derivatives?
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Given an parabola $\displaystyle y=ax^2+bx+c,a\not = 0$ and a line $\displaystyle y=px+q$. This line is tangent to the parabola if and only if $\displaystyle (b-p)^2 - 4a(c-q)=0$. So if $\displaystyle px+q$ is tangent to $\displaystyle x^2$ then it means $\displaystyle p^2 + 4q = 0$ and it tangent to $\displaystyle -x^2+2x-5$ that means $\displaystyle (p-2)^2 - 4(q+5)=0$. The solutions to this system of equations is $\displaystyle (p,q) = \{(-2,1),(4,-4)\}$.
    the other solution to the system is (-2,-1), not (-2,1)
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  5. #5
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    Quote Originally Posted by doctorgk View Post
    how did you get that formula? is there a way to solve this with derivatives?
    Given a parabola $\displaystyle y=ax^2+bx+c,a\not = 0$ and $\displaystyle y=px+q$ if they intersect exactly once then the equation (of the intersection) $\displaystyle ax^2+bx+c = px+q$ has exactly one solution, this is a quadradic so, $\displaystyle ax^2+(b-p)x+(c-q)=0$. We want this quadradic to has exactly one real solution that happens when the discrimant is zero.

    (We want it to have exactly one solution because a tangent line to a parabola intersects it exactly one time. )

    the other solution to the system is (-2,-1), not (-2,1)
    Okay I fix it.
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  6. #6
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    Okay

    makes sense now, thank you very much
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