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Math Help - Another Tangent line question

  1. #1
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    Another Tangent line question

    Consider the curve defined by x^2 + 4y^2 = 7 + 3xy

    Show that there is a point P with x-coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P

    dy/dx = (3y - 2x)/ (8y-3x)
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ny_chow View Post
    Consider the curve defined by x^2 + 4y^2 = 7 + 3xy

    Show that there is a point P with x-coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P

    dy/dx = (3y - 2x)/ (8y-3x)
    I think you forgot to tell us what the x-coordinate is. For now I'll just call it x_0.

    So the derivative is \frac{dy}{dx}=\frac{3y-2x}{8y-3x}. The horizontal tangent line occurs when the derivative is zero, namely, when 3y-2x=0. Since we are given the x-coordinate (which I'm calling x_0), we have that 3y-2x_0=0. Solving for y gives us y=\frac{2x_0}{3} so the point is P=\left(x_0,\frac{2x_0}{3}\right).

    So whatever the x-coordinate actually is, just plug it in for x_0 and that will give you the answer.
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  3. #3
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    oh right the x coordinate is 3.

    yeah thats what i did, but the answer key says to reject y = 2,
    and that the answer is 0.165
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  4. #4
    Super Member redsoxfan325's Avatar
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    Are you sure you're looking at the right answer? Because y=2 is the only horizontal tangent line to f(x) at x=3. I graphed it to be sure. (It's an ellipse, by the way)
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  5. #5
    Super Member redsoxfan325's Avatar
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    To expand a bit on my above answer, it's an ellipse rotated 22.5^o above the x-axis.
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