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Math Help - Two tangent Lines to an ellipse

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    Two tangent Lines to an ellipse

    From the point (14,14) there are two tangent lines to the ellipse 3x^2+y^2=28. Find the coordinates of the two points of tangency.

    So I started by using implicit differentiation to find dy/dx = -3x/y and we know that y = sqrt(28-3x^2) but I'm not sure what to do from here. Would I plug in y so that dy/dx = -3x/sqrt(28-3x^2)? If that is what you do then how does that help you find the coordinates?

    Thanks in advance for help
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    Quote Originally Posted by Alice96 View Post
    From the point (14,14) there are two tangent lines to the ellipse 3x^2+y^2=28. Find the coordinates of the two points of tangency.

    So I started by using implicit differentiation to find dy/dx = -3x/y and we know that y = sqrt(28-3x^2) but I'm not sure what to do from here. Would I plug in y so that dy/dx = -3x/sqrt(28-3x^2)? If that is what you do then how does that help you find the coordinates?

    Thanks in advance for help
    Yes, use the derivative \frac{dy}{dx} = - \frac{3x}{y}. Suppose the two tangent points are (x_1,y_1) \; \text{and}\; (x_2,y_2). Then the equation of each tangent is

     <br />
y - y_1 = -\frac{3x_1}{y_1} \left(x - x_1\right),\;\;\; <br />
y - y_2 = -\frac{3x_2}{y_2} \left(x - x_2\right)<br />

    Now each tangent line passes throught (14,14) so 14 - y_1 = -\frac{3x_1}{y_1} \left(14 - x_1\right) or simplifying gives 14 y_1 + 3 \cdot 14 x_1 = 3x^1_2 + y_1^2 = 28 or y_1 + 3x_1 = 2. Now eliminate y_1 from 3x_1^2 + y_1^2 = 28 giving a single quadratic equation for x_1 (which you can solve). Using y_1 + 3x_1 = 2 gives you the y value.
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