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Math Help - use dy/dx to find the equ. of tangent

  1. #1
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    use dy/dx to find the equ. of tangent

    bk ex18a p61 q28
    find the equations of the two tangents to the curve y^2 = x^2 y + 2 at the point where x = 1

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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    bk ex18a p61 q28
    find the equations of the two tangents to the curve y^2 = x^2 y + 2 at the point where x = 1

    Hello,

    the "b" of the equation of the tangent is the y-coordinate of the tangent point. The tangent point must be situated on the given curve. Therefore calculate first the coordinates of the tangent point.

    You know already that the x-coordinate of the tangent point is x = 1. Plug in this value into the given equation:

    y^2 = 1^2 \cdot y + 2~\implies~y^2-y-2=0~\implies~y=2~\vee~y=-1

    That means you have 2 points which satisfy the conditions: T_1(1, 2), T_2(1, -1)

    Plug in the values of the coordinates into the equation of the dy/dx to calculate the slope of the tangents.
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  3. #3
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    Hello,

    in addition to my previous post I've attached a drawing of the tangents and the curve.
    Attached Thumbnails Attached Thumbnails use dy/dx to find the equ. of tangent-implizit_tangenten.gif  
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  4. #4
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    Quote Originally Posted by afeasfaerw23231233 View Post
    bk ex18a p61 q28
    find the equations of the two tangents to the curve y^2 = x^2 y + 2 at the point where x = 1

    Two tangent lines where x = 1?

    That means there are two ponts on the curve where x=1.

    y^2 = (x^2)y +2
    y^2 = (1^2)y +2
    y^2 = y +2
    y^2 -y -2 = 0
    (y -2)(y +1) = 0
    y = 2 or -1
    So points (1,2) and (1,-1)

    You got the dy/dx, which is the slope of the tangent line anywhere on the curve.

    dy/dx = 2xy / (2y -x^2)

    ------------------------------------
    At point (1,2)

    slope, m1 = (2*1*2) / (2*2 -1^2) = 4/3.

    So we have a point and the slope.
    Hence, use the point-slope form of the equation of a line,
    (y -y1) = m(x -x1)

    (y -2) = (4/3)(x -1)
    y -2 = (4/3)x -4/3
    y = (4/3)x -4/3 +2
    y = (4/3)x +(-4 +3*2)/3
    y = (4/3)x +2/3 ---------------------one tangent line.

    -----------------------------------------
    At point (1,-1)

    m2 = (2*1*(-1)) / (2(-1) -1^2) = -2/(-3) = 2/3

    y -(-1) = (2/3)(x -1)
    y +1 = (2/3)x -2/3
    y = (2/3)x -2/3 -1
    y = (2/3)x -5/3 --------------the other tangent line.
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  5. #5
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    thanks a lot. i misunderstood the question. i thought the the point (1,b) is a point on the tangent which lies outside the circle. but if it is the case, can i find the equation of the tangent?
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  6. #6
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    Quote Originally Posted by afeasfaerw23231233 View Post
    thanks a lot. i misunderstood the question. i thought the the point (1,b) is a point on the tangent which lies outside the circle. but if it is the case, can i find the equation of the tangent?
    Hello,

    which circle?
    The graph of the given equation consists of 2 seprate parts but it dosn't describe a circle.
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