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Math Help - Determine is a line is tangent to a curve or not?

  1. #1
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    Determine is a line is tangent to a curve or not?

    Sorry if it sounds silly but I am thinking if there is a way to find out if a line is a tanget of a curve?

    Say given a curve of x^2 + 3xy + y^2 + 4 = 0, how do I determine if any one given line is a tangent of this curve or not? The line can be y=x+7 or y=-2x-4, etc. And if it is a tangent of the curve, how do I find the points that this line is tangent to the curve?

    I am thinking that I differentiate the equation of the curve to have dy/dx. Then I can get the gradient at any one point of the curve to compare with a line. But this isn't enough because same gradient doesn't mean is a tangent to the curve. And how I find the points which the line is tangent to the curve without any given points at all?

    Thanks for any help!
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Do a simultaneous equation to determine first whether the line meet the curve or not.

    Once done, you need to find if the gradient at the point of intersection is the same as the gradient of the line.

    If they meet and share a common gradient, you have your tangent. If one of those two conditions are not met, the line is not a gradient.
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  3. #3
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    q.[/math]

    Hello, xEnOn!

    I have a method but it's quite long.
    Maybe someone has a shorter approach?


    Is there a way to find out if a line is a tanget to a curve?

    Say, given a curve: . x^2 + 3xy + y^2 + 4 \:=\: 0
    how do I determine if, say, y \:=\:x+6 is a tangent of this curve or not?
    And if it is a tangent of the curve, how do I find the points of tangency?

    I am thinking that I differentiate the equation of the curve to have dy/dx.
    Then I can get the gradient at any one point of the curve to compare with the line.
    But this isn't enough because same gradient doesn't mean it is tangent to the curve.
    And how I find the points which the line is tangent without any given points at all?

    We have: . \begin{Bmatrix}x^2 + 3xy + y^2 + 4 \;=\; 0 & [1] \\ y \;=\; x + 6 & [2] \end{Bmatrix}


    Find the points of intersection, substitute [2] into [1]:

    . . x^2 + 3x(x+6) + (x+6)^2 + 4 \:=\:0

    . . x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \:=\:0

    . . 5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0

    . . (x+2)(x+4)\:=\:0 \quad\Rightarrow\quad x \:=\:-2,\:-4

    Substitute into [2]: . y \:=\:4,\:2


    The hyperbola and the line intersect at: . P(-2,4)\,\text{ and }\,Q(-4,2)


    If the line is tangent to the hyperbola, their slopes will be equal at \,P and Q.


    The slope of the line is: . \boxed{m \,=\,1}


    Find the slope of the hyperbola.

    Differentiate implicitly: . 2x + 3x\dfrac{dy}{dx} + 3y + 2y\dfrac{dy}{dx} \:=\:0

    . . (3x + 2y)\dfrac{dy}{dx} \:=\:-2(x+y) \quad\Rightarrow\quad \boxed{\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(x+y)}{3x+2y} }



    \text{At }P(\text{-}2,4)\!:\;\;\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(\text{-}2+4)}{3(\text{-}2) + 2(4)} \:=\:\dfrac{\text{-}4}{2} \:=\:-2 \;\ne \;1


    \text{At }Q(\text{-}4,2)\!:\;\;\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(\text{-}4+2)}{3(\text{-}4)+2(2)} \:=\:\dfrac{4}{\text{-}8} \:=\:-\dfrac{1}{2} \;\ne\;1


    The slopes are not equal at the intersection points.

    Therefore, the line is not tangent to the hyperbola.



    Edit: Too slow again!
    . . . .Unknown008 already gave an excellent game plan.
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  4. #4
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    Quote Originally Posted by Soroban View Post
    I have a method but it's quite long.
    Maybe someone has a shorter approach?

    We have: . \begin{Bmatrix}x^2 + 3xy + y^2 + 4 \;=\; 0 & [1] \\ y \;=\; x + 6 & [2] \end{Bmatrix}


    Find the points of intersection, substitute [2] into [1]:

    . . x^2 + 3x(x+6) + (x+6)^2 + 4 \:=\:0

    . . x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \:=\:0

    . . 5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0

    . . (x+2)(x+4)\:=\:0 \quad\Rightarrow\quad x \:=\:-2,\:-4
    A somewhat shorter method is to follow Soroban's approach by substituting the line equation into the conic equation, as above, and finding the values of x where the line meets the conic. These turned out to be –2 and –4. Those values are different, so you can tell straight away that the line cannot be a tangent to the conic. If it were, then there would be a repeated root for x (because the point at which the line touches the conic would be a double root of the equation). As it is, the line crosses the conic at two distinct points, so it cannot touch it.

    Suppose that we have the conic equation x^2 + 3xy + y^2 + 4 = 0, as before, but that the equation of the line is 4x+y+4=0. Substitute y = -4x-4 into the conic equation, as in Soroban's method. You get
    x^2 + 3x(-4x-4) + (-4x-4)^2 + 4 = 0,
    x^2 - 12x^2 - 12 + 16x^2 + 32x + 16 + 4 = 0,
    5x^2 + 20x + 20 = 0,
    x^2+4x+4=0,
    (x+2)^2 = 0.

    This time, there is a repeated root x = –2, which tells you that the line touches the conic at that value of x. So in that case, the line is tangent to the conic.
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  5. #5
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    By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?

    I suppose I will still get a result in someway no matter when I do simultaneous?

    Also, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation.
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  6. #6
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    Quote Originally Posted by xEnOn View Post
    By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?
    If you're talking about binomials, (x+a)(x+b): If the simultaneous equation gives 0 or 2 answers, then it is not tangent.
    In other cases, it is quite lengthy - you have to find the co-ordinates of where they meet and then check if their gradients are the same for both equations at that point.
    You cannot easily determine whether it is tangent or not, just from the intersection points. (Unless there are no intersection points)

    Quote Originally Posted by xEnOn View Post
    I suppose I will still get a result in someway no matter when I do simultaneous?
    No. Well not real results anyway (but who wants to get into imaginary numbers?). Sometimes when you do simultaneous equations, you won't get any answers and this will tell you straight away that the 2 graphs don't meet.

    Quote Originally Posted by xEnOn View Post
    Also, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation.
    I'm not getting your question...
    Both equations only have the 2 unknown variables in it.

    What do you mean you take the wrong equation to do a simultaneous equation?
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