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Math Help - Finding equation of tangent lines:

  1. #1
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    Finding equation of tangent lines:

    Hello,
    So I am having problems with trying to find the equations of tangent lines when the point the lines pass through is not on the original function.

    ex:
    f(x)=x^2
    find equations of tangent lines through the point (1,-4)
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  2. #2
    Member sbhatnagar's Avatar
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    Lightbulb Re: Finding equation of tangent lines:

    we need to find the tangents to the curve y=x^2 passing through (1,-4).

    y=x^2

    \Rightarrow \frac{dy}{dx}=2x ...(i)

    (1)Let the tangent pass through the point (a,b) such that (a,b) lies on y=x^2, then the slope of the tangent is m=2a(using (i)). Since (1,-4) lies on the tangent, the equation of the tangent is y=2ax-a^2 or y=2ax-(2a+4).

    (2)Also note that slope, m=\frac{b+4}{a-1}=\frac{a^2+4}{a-1}. We also know that m=2a (we found it in point 1 using differentiation).

    Relating these two: 2a=\frac{a^2+4}{a-1}
    \Rightarrow a^2-2a-4=0.
    Solving this with quadratic formula : a=1 \pm \sqrt{5}. Since there are two values of a, there must be two equations.

    (3)Put the values of a in any of the two equations (in point 1) and you will get: y=(2+2\sqrt{5})x-(6+2\sqrt{5}) and y=(2-2\sqrt{5})x+(-6+2\sqrt{5}).

    Last edited by sbhatnagar; October 15th 2011 at 04:38 AM.
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  3. #3
    Senior Member DeMath's Avatar
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    Re: Finding equation of tangent lines:

    As we know, the tangent equation is the straight line y=ax+b.

    From this "...through the point (1,-4)" we get the condition a+b=-4 (why?).

    Also, the equations y=x^2,~y=ax+b must have one common point; therefore, the quadratic equation x^2=ax+b must have a single root, which is possible, if his discriminant is equal to zero \Delta=a^2+4b=0.

    So, the unknown parameters of the tangent equations is solutions of this equations system

    \begin{cases}a+b=-4,\\ a^2+4b=0.\end{cases} which has two pairs of roots \begin{cases}a_{1,2}=2\pm2\sqrt{5},\\ b_{1,2}=-6\mp2\sqrt{5}.\end{cases}

    Finally, we have y=(2+2\sqrt{5})x+(-6-2\sqrt{5}) and y=(2-2\sqrt{5})x+(-6+2\sqrt{5}).

    See the picture

    Last edited by DeMath; October 15th 2011 at 04:05 AM. Reason: typo
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  4. #4
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    Re: Finding equation of tangent lines:

    As long as f is a polynomial, you can solve an problem like this using Fermat's method of "ad-equation" (predating Calculus). As has been said, the equation of any line through (1, -4) must be of the form y= m(x- 1)- 4. Also as has been said, at a point of tangency, the curve and line must cross: y= x^2= m(x- 1)- 4 which gives the quadratic equation y^2- mx+ (m+ 4)= 0.

    But in order to be tangent, that x must be a double root of the equation. And that means the "discriminant", b^2- 4ac= (m^2)- 4(1)(m+4)= 0. That gives a quadratic equation for m which has two roots as sbhatnagar and deMath indicate.
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