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Math Help - Equations of tangent lines of an ellipse

  1. #1
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    Equations of tangent lines of an ellipse

    How do I solve this question?

    Find the equations of both the tangent lines to the ellipse x^2 + 4y^2 = 36 that pass through the point (12, 3).

    There is supposed to be 2 tangent lines, a horizontal and non-horizontal one.

    Thank you in advance!
    Last edited by zer0e; October 6th 2009 at 07:23 PM.
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  2. #2
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    Quote Originally Posted by zer0e View Post
    How do I solve this question?

    Find the equations of both the tangent lines to the ellipse x2 + 4y2 = 36 that pass through the point (12, 3).

    There is supposed to be 2 tangent lines, a horizontal and non-horizontal one.

    Thank you in advance!
    x^2+4y^2=36

    The second you hear tangent line, think derivative. Since this has a y^2 in it, we're going to differentiate implicity

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

    Moving stuff around,

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


    So the slopes of our lines must have the form above to be tangent

    Now we have 1 point, so let's find the toher point

    Points on the elipse are represented by (x,\sqrt{\frac{36-x^2}{4}}) or (x,-\sqrt{\frac{36-x^2}{4}})


    And slope= \frac{y_1-y_2}{x_1-x_2}

    Sooooo,

    The derivative equation must equal the slope between the 2 points and lets use our formulas to plug in for y

    -\frac{x}{4\sqrt{\frac{36-x^2}{4}}}=\frac{3-\sqrt{\frac{36-x^2}{4}}}{12-x}

    And don't forget that the y values can be represented 2 ways (+ and -)

    -\frac{x}{4\cdot -\sqrt{\frac{36-x^2}{4}}}=\frac{3+\sqrt{\frac{36-x^2}{4}}}{12-x}

    Solving both of those will get you your x values, and then to get your y's remember to keep to + and -'s straight

    Then the equation of the tangent lines will be

    y-y_0=-\frac{x}{4y}(x-x_0)

    Of course, you already said 1 tangent line will be horizontal, and you know the point it goes through, so you can get that equation almost without thinking
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