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Math Help - One Last question about ellipse

  1. #1
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    One Last question about ellipse

    The perimeter of a triangle is 30 and the points (0,5) & (0, -5) are two of the vertices. Find the graph of the third vertex.

    Answer - \frac{x^2}{75} + \frac{y^2}{100} = 1

    Thanks very much for all the help
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    The perimeter of a triangle is 30 and the points (0,5) & (0, -5) are two of the vertices. Find the graph of the third vertex.

    Answer - \frac{x^2}{75} + \frac{y^2}{100} = 1

    Thanks very much for all the help
    Last question on ellipses? You got tired of them?

    Imagine, or plot those two given points on an x,y set of coordinates. So one side of the triangle is 10 units long. The other two sides should be 30-10 = 20units combined.

    Now we play. Place the 3rd corner along the y-axis. No triangles here. But the 3rd "corner" is 5 units above "corner" (0,5) or 5 units below "corner" (0,-5). So the vertical axis, 2b, of the ellipse is 20 units long.

    Place the 3rd corner on the, say, positive x-axis. We have an isosceles triangle, made up of two congruent right triangles.
    In any of the right triangles: hypotenuse = 20/2 = 10 units. Vertical leg = 5 units. So horizontal leg = sqrt(10^2 -5^2) = sqrt(75) units.
    Meaning, the horizontal axis, 2a, of the ellipse is 2sqrt(75) units long.

    The ellipse:
    (x^2)/(a^2) +(y^2)/(b^2) = 1
    (x^2)/(sqrt(75))^2 +(y^2)/(10^2) = 1
    (x^2)/75 +(y^2)/100 = 1 ------------------------answer.
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  3. #3
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    Hello, ^_^Engineer_Adam^_^!

    The perimeter of a triangle is 30 and the points (0,5) & (0, -5) are two of the vertices.
    Find the graph of the third vertex.

    Answer: \frac{x^2}{75} + \frac{y^2}{100} \:= \:1

    I won't make a sketch this time.


    We have vertices: .A(0,5), B(0,-5), C(x,y).


    The perimeter is 30: .AB + BC + CA .= .30

    Since AB = 10, we have: .BC + CA .= .20

    . . . . . . . . . . . . . . . . . . . . . . . . . . . ____________ . . . . . . . ___________
    Using the Distance Formula: . BC .= .√(x + (y + 5), . CA .= .√(x + (y - 5)

    . . . . . . . . . ___________ . . .___________
    We have: . √x + (y + 5) + √x + (y - 5) . = . 20
    . . . . . . . . . . . . . . . . . . . . . ___________ . . . . . . . __________
    . . . . . . . . . . . . . . . . . . . . √x + (y + 5) .= .20 - √x + (y - 5)

    . . . . . . . . . . . . . . . . . . . . . . . . . ___________
    Square both sides and simplify: . 2√x + (y + 5) . = . y + 20


    Square both sides and simplify: . 4x + 3y .= .300

    . . . . . . . . . . . . x . . .y
    Divide by 300: . ---- + ---- . = . 1
    . . . . . . . . . . . .75 . . 100

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