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Math Help - Tangent Passing Through a Point & Implicit Differentiation

  1. #1
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    Tangent Passing Through a Point & Implicit Differentiation

    Question:
    Find the equation(s) of the tangents to x^2 + 4y^2 = 16 which pass through
    a) the point (2,2)
    b) the point (4,6).

    Attempted Solution:
    y = \pm\sqrt{\frac{-x^2 + 16}{4}}

    y' = \frac{-x}{4y}

    a) For P(2,2):

    Let P_{tan} be parametrized as (a, f(a).

    \frac{2 - \pm\sqrt{\frac{-a^2 + 16}{4}}}{2 - a} = \frac{rise}{run}= m

    What do I do from here? The points given are not points of tangency; they are points that the tangents pass through. So, do I have to set something equal and solve for a?
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  2. #2
    Senior Member Sambit's Avatar
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    \frac{y-2}{x-2}=[\frac{dy}{dx}]_{(x,y)=(2,2)}
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    Do I have to sub P(2,2) into the derivative? What exactly am I trying to solve for?
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  4. #4
    Senior Member Sambit's Avatar
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    after gettin \frac{dy}{dx}, substitute x by 2 and y by 2.
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    That leaves me with a slope of 1/4. Do I set 1/4 equal to the rise/run equation in my OP?
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  6. #6
    Senior Member Sambit's Avatar
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    That leaves you with the equation \frac{y-2}{x-2}=1/4
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  7. #7
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    Quote Originally Posted by Sambit View Post
    \frac{y-2}{x-2}=[\frac{dy}{dx}]_{(x,y)=(2,2)}
    We need to take care as the point (2,2) is not on the ellipse.

    If you plot the question both answers can be read off the graph. The tangent lines must be horizontal and vertical. Note that

    \displaystyle x^2+4y^2=16 \iff \frac{x^2}{4^2}+\frac{y^2}{2^2}=1

    Is the equation in standard form.

    Tangent Passing Through a Point & Implicit Differentiation-ellipse.jpg



    To solve it algebraically we need to find a point (a,b) on the ellipse e.g it must satisfy a^2+4b^2=16 and

    \frac{b-2}{a-2}=\frac{dy}{dx}\bigg|(a,b)=-\frac{a}{4b}

    Multiplying out the bottom equation gives

    4b^2+a^2=2a+8b but since it must be on the ellipse  a^2+4b^2=16 this gives

    16=2a+8b \implies a=8-4b=4(2-b)

    Now putting this back into the ellipse again gives

    [4(2-b)]^2+4b^2=16 \iff 5b^2-16b+12=0 \iff (5b-6)(b-2)=0 as you can check this gives b=2 \implies a=0 and correctly gives the point (0,2) as when one of the tangent lines occur.
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