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Math Help - Another ellipse question

  1. #1
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    Another ellipse question

    Stuck on my homework question...again lol...

    The tangents of the ellipse with equation x^2/(a^2)+y^2/(b^2)=1 at the points P(acost, bsint) and Q(-asint, bcost) intersect at the point R. (I tried to find R by calculating the equation of the two tangents and then equating them to obtain the x and y coordinates). As t varies, show that R lies on the curve with equation x^2/(a^2)+y^2/(b^2)=2. I then tried to eliminate t to get an equation with just x's and y's but I couldn't eliminate t, so I must have gone wrong. When I tried to get the x-coordinate of R, everything cancelled to 0.

    Could someone help please?
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  2. #2
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    Why try to eliminate the parameter? There is a reason why folks got into parametric equations. Sometimes, you can do more things.

    Find the general slope.

    \frac{dy}{dx}\;=\;-\frac{b^{2}x}{a^{2}y}

    You have to know where that came from.

    Now, write the equations of the tangent lines.

    y-b\sin(t)\;=\;\frac{dy}{dx}(x-a\cos(t))

    y-b\cos(t)\;=\;\frac{dy}{dx}(x+a\sin(t))

    Don't forget to substitute the right values in dy/dx.

    Solve system, for x and y.

    I get x = a(cos(t)-sin(t)) and y = b(cos(t)+sin(t))

    One last thing. Is that point on the new ellipse?

    This is a great review problem. There is a ton of mathematics in there. Understand this problem and you will have some good information for the future.
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  3. #3
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    Hello, eveyone!


    When working with tangents to conic curves,
    . . there are some formulas that can save a lot of time and work.


    Given the point P(x_1,y_1) on the circle: . x^2 + y^2 \:=\:r^2

    . . the equation of the tangent at P is: . x_1x + y_1y \:=\:r^2


    Given the point P(x_1,y_1) on the ellipse: . \frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1

    . . the equation of the tangent at P is: . \frac{x_1x}{a^2} + \frac{y_1y}{b^2} \:=\:1


    Given the point P(x_1,y_1) on the hyperbola: . \frac{x^2}{a^2} - \frac{y^2}{b^2} \:=\:1

    . . the equation of the tangent at P is: . \frac{x_1x}{a^2} - \frac{y_1y}{b^2} \:=\:1

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