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Math Help - Gradient of the tangent to the ellipse

  1. #1
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    Gradient of the tangent to the ellipse

    I am stuck on this question.

    x = root 3 cos t, y = sin t (t\in [0,2pi])

    point P is where t = 5pi/6, ie (-1.5,0.5)

    The gradient of the tangent to the ellipse at the point with parameter t is -cos t/root 3 sin t, where t is not equal to 0, pi, 2 pi.
    I need to find the equation of the tangent to the ellipse at the point P.

    Can anyone help?
    Last edited by Arron; April 27th 2011 at 11:41 AM.
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  2. #2
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    Quote Originally Posted by Arron View Post
    I am stuck on this question.
    x = root 3 cos t, y = sin t (t\in [0,2pie])
    point P is where t = 5pie/6, ie (-1.5,0.5)
    Can you find each of these
    The evaluate at the given point.
    How do you use those to find the gradient?

    BTW: \pi is named pi not pie.
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  3. #3
    Senior Member Sambit's Avatar
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    The gradient at point with parameter t is -cos t/root 3 sin t; and the point P has parameter t=5pi/6. So the gradient of the tangent will be -cos(5pi/6) / root (3 sin (5pi/6)) ie. 1/sqrt(2)

    Again the co-ordinates of point P are (-1.5 , .5). So using Point Slope Form of equation of straight line, the tangent's equation will be (y-0.5)/(x-(-1.5)) = (1/sqrt(2))

    EDIT:
    \pi is named pi not pie
    That was really cool
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  4. #4
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    Thank you so much for your help. Finally could you please let me know how to find the point where the tangent meets the x-axis.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Arron View Post
    Thank you so much for your help. Finally could you please let me know how to find the point where the tangent meets the x-axis.
    You have the equation of the line, when the line reaches the x axis the value of y is zero. Solve for x.

    -Dan
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  6. #6
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    Quote Originally Posted by Arron View Post
    Thank you so much for your help. Finally could you please let me know how to find the point where the tangent meets the x-axis.
    BUT be careful.
    The answer in reply #3 is not correct.
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  7. #7
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Plato View Post
    BUT be careful.
    The answer in reply #3 is not correct.
    Why?
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  8. #8
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    Quote Originally Posted by Sambit View Post
    Why?
    The slope is not correct.
    What is \dfrac{dy}{dx}\left(\frac{5\pi}{6}\right)~?
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  9. #9
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    The denominator should be -root(3) sin(5pi/6), not root (3 sin (5pi/6))
    And, of course, the derivative of sin(t) is cos(t) not -cos(t) but sambit may have just moved the - from the denominator to the numerator,
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  10. #10
    Senior Member Sambit's Avatar
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    Ok I got it. Sorry for that.
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  11. #11
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    Guys are you saying that the correct answer is cos(5pi/6)/-Root 3 sin (5pi/6) = 1/Root 2.
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  12. #12
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    I am still not sure about the gradient, am I correct in the last post?
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