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Math Help - elliptical question

  1. #1
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    Smile elliptical question

    hi guys,

    I have the following problem:
    The point P lies on the ellipse x^2 + 4y^2 = 1 and N is the foot of the perpendicular from P to the line x = 2 (see attachment). Find the equation of the mid point of PN as P moves on the ellipse.

    My answer does not agree with the one in my maths book, so I would just like to confirm whether the book is correct or not. Secondly, I have solved this problem using the parametric form of the point (x = a.cos \theta, y= a.sin \theta) and as far as I can see, it is the only way to solve this problem - but I thought the parametric form was simply an alternate formulation of the ellipse and therefore there should be a way of solving the question without using the parametric form. Is this right?

    So comparing x^2 + 4y^2 = 1 with \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 we have a = 1 and b = \frac{1}{2}

    P(x = a.cos \theta, y= a.sin \theta) and N(2,y) so the mid point is M(\frac{a.cos \theta + 2)}{2}, a.sin \theta)

    therefore 2x = a cos \theta + 2
    4x^2 = (a.cos \theta + 2)^2

    substitute:
    y = a sin \theta
    2x - 2 = a. cos \theta

    I get: 4x^2 + y^2 - 8x + 3 = 0
    My maths book gives: 4x^2 + 4y^2 - 8x + 3 = 0

    regards
    Attached Thumbnails Attached Thumbnails elliptical question-m16g14.jpg  
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  2. #2
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    Quote Originally Posted by s_ingram View Post
    hi guys,

    I have the following problem:
    The point P lies on the ellipse x^2 + 4y^2 = 1 and N is the foot of the perpendicular from P to the line x = 2 (see attachment). Find the equation of the mid point of PN as P moves on the ellipse.

    My answer does not agree with the one in my maths book, so I would just like to confirm whether the book is correct or not. Secondly, I have solved this problem using the parametric form of the point (x = a.cos \theta, y= a.sin \theta) and as far as I can see, it is the only way to solve this problem - <<<<<< that is correct but the parametric equation of an ellipse is: (x = a\cdot cos \theta, y= \bold{\color{red}b} \cdot sin \theta).
    If you use this equation and the values for a and b you already know you'll get the equation which is given by your textbook.

    The equation describes a semi-circle around (1, 0) with radius r = 0.5.
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  3. #3
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    Yes, silly mistake. Thanks for finding it. But why do you say the locus is a semi circle? Isn't it a circle?
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  4. #4
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    Quote Originally Posted by s_ingram View Post
    Yes, silly mistake. Thanks for finding it. But why do you say the locus is a semi circle? Isn't it a circle?
    Correct, you are right.
    I split the point P into 2 points A and B to construct the locus: The point D (in red) is the midpoint of AN and the point C (in blue) is the midpoint of BN.

    Then you'll get two semi-circles which form a complete circle.
    Attached Thumbnails Attached Thumbnails elliptical question-locus_elli_krs.png  
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