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Math Help - Please help!

  1. #1
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    Smile Please help!

    The graphs of x/4 + y/3 = 1 and (x^2)/16 + (y^2)/9 = 1 intersect at points A and B. Find the coordinates of a third point C on the graph of the second curve such that the area of triangle ABC is equal to 3.



    Thank you very much for your help!! I appreciate it!
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  2. #2
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    Quote Originally Posted by rlarach View Post
    The graphs of x/4 + y/3 = 1 and (x^2)/16 + (y^2)/9 = 1 intersect at points A and B. Find the coordinates of a third point C on the graph of the second curve such that the area of triangle ABC is equal to 3.



    Thank you very much for your help!! I appreciate it!
    Since no-one else has replied I'll suggest an approach:

    1. Go to parametric equation for the ellipse: x = 4 cos t, y = 3 sin t. Then the vertices of the triangle are A(0, 3), B(4, 0) and C(4 cos t, 3 sin t).

    2. The area of a triangle whose vertices have coordinates (x1, y1), (x2, y2) and (x3, y3) is A = 1/2 |(x1 - x3)(y1- y3) - (y1 - y2)(x1 - x3)| (you should prove this - only takes a couple of lines).

    3. Sub in the coordinates of the vertices of the triangle and solve A = 3. You get a trig equation that you may need help solving.

    There's probably a much more elegant approach - maybe it's no coincidence that if you consider the point D(0, -3) then triangle ABD has an area of 12 = 4 x 3 .....
    Last edited by mr fantastic; January 3rd 2008 at 12:09 AM. Reason: Forgot the 1/2 (in red)
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mr fantastic View Post
    There's probably a much more elegant approach
    Actually, since I tend to "bull" my way through a problem, I thought your solution was quite elegant.

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    Actually, since I tend to "bull" my way through a problem, I thought your solution was quite elegant.

    -Dan
    Ta. Actually I tend to bull$&# my way through a problem

    (You know what they say .... If you can't blind them with you brilliance, baffle them with you bull$&#
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by mr fantastic View Post
    Ta. Actually I tend to bull$&# my way through a problem

    (You know what they say .... If you can't blind them with you brilliance, baffle them with you bull$&#
    I had to thank you for saying one of my favorite lines!

    "If you can't blind them with you brilliance, baffle them with you bull$&#" -Anonymous
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mr fantastic View Post
    (You know what they say .... If you can't blind them with you brilliance, baffle them with you bull$&#
    Hmmmm... Sounds like you attended my Thesis presentation for my MS.

    -Dan
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  7. #7
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    Quote Originally Posted by rlarach View Post
    The graphs of x/4 + y/3 = 1 and (x^2)/16 + (y^2)/9 = 1 intersect at points A and B. Find the coordinates of a third point C on the graph of the second curve such that the area of triangle ABC is equal to 3.
    Thank you very much for your help!! I appreciate it!
    There is another approach(a little easier in finding the area of the triangle)

    First off, Mr.F's (1) says (4,0) and (0,3) are the points...

    Now I assume you know the perpendicular distance of a point (m,n) from the line Ax+By+C = 0 is \frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}----(*)

    From this it is a tad easier because AB = \sqrt{(3-0)^2+(0-4)^2} = 5
    From
    and height(from (*)) is \frac{|\frac{m}{4}+\frac{n}{3}-1|}{\sqrt{(\frac14)^2+(\frac13)^2}}
    and (m,n) lies on the ellipse ,so (m,n) = (4 \cos t, 3 \sin t)
    Now, Area = \frac12 \cdot base \cdot height
    Base = AB = 5
    Height = \frac{|\frac{4 \sin t}{4}+\frac{3\cos t}{3}-1|}{\sqrt{(\frac14)^2+(\frac13)^2}}= \frac{12}{5} \cdot |\sin t + \cos t - 1|
    You want A to be 3, which means
    |\sin t + \cos t - 1| = \frac12
    From now on, it is just basic trigonometry
    This is not less messier than Mr.Fs way, but a different approach
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