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Math Help - differentiation question 2

  1. #1
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    Exclamation differentiation question 2

    Find the equations of the lines that are tangents to the elipse
    x^2 + 4y^2 = 16 and that also passes through the point (4,6).
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  2. #2
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    Hi

    First the line x=4 is tangent to the ellipse x^2 + 4y^2 = 16 and passes through the point (4,6).

    If there is another one its equation is y=mx+p
    It is tangent to the ellipse. Therefore there is only one contact point between the ellipse and the line. In other words the system
    x^2 + 4y^2 = 16
    y=mx+p
    has one single solution

    It means that the equation x^2 + 4(mx+p)^2 = 16 or (1+4m^2)x^2+8mpx+4p^2-16=0 has one solution. Its discriminant is thus equal to 0.
    (4mp)^2-(1+4m^2)(4p^2-16)=0
    -p^2+4+16m^2=0

    The line passes through the point (4,6). Therefore 6=4m+p or p=6-4m.

    Then -(6-4m)^2+4+16m^2=0 giving m=\frac{2}{3} and p=\frac{10}{3}
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  3. #3
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    Quote Originally Posted by math123456 View Post
    Find the equations of the lines that are tangents to the elipse
    x^2 + 4y^2 = 16 and that also passes through the point (4,6).
    Here is the idea. Lets suppose that the point of tangency is (a,b)

    Plugging this point in we get our 1st equation

    E_1: a^2+4b^2=16

    Now lets take the derivative of the equation(implicitly)

    2x+8y\frac{dy}{dx}=0 \iff \frac{dy}{dx}=\frac{-x}{4y}

    So now we know the slope is \frac{dy}{dx}=\frac{-a}{4b}

    We also know that the tangent line passes through (4,6) so we get the slope is

    m=\frac{6-b}{4-a} and we know that the slopes have to be equal.... so setting them equal yields our 2nd equation.

    E_2: \frac{6-b}{4-a}=\frac{-a}{4b}

    This is a system of non-linear equations...

    Solveing yeilds the solution \left(\frac{-16}{5},\frac{6}{5}\right)

    So plugging into the equation for the derivative we get

    \frac{dy}{dx}=\frac{\frac{16}{5}}{(4)\frac{6}{5}}=  \frac{2}{3}

    y=\frac{2}{3}x+b \implies 6=\frac{2}{3}(4)+b \iff \frac{10}{3}=b

    So we end up with y=\frac{2}{3}x+\frac{10}{3}
    Good luck.

    TES
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