Calculus

Convert the polar equation to rectangular coordinates: r = 2-cos theta

...and find all points of vertical and horizontal tangency.

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  1. we know x^2 + y^2 = r^2
    and cosØ = x/r
    so, ...
    r = 2-cos theta
    r = 2 - x/r
    r^2 = 2 - x
    x^2 + y^2 = 2-x

    2x + 2y d/dx = -1
    2y dy/dx = -1 - 2x
    dy/dx = (-1-2x)/(2y
    = 0 for a horizontal tangent
    -1 - 2x = 0
    x = -1/2
    and subbing back into x^2 + y^2 = 2-x
    1 + y^2 = 2 + 1
    y^2 = 2
    y = ± √2

    there are 2 horizontal tangents
    at ((-1/2) , √2) and (-1/2 , - √2)

    for a vertical tangents, dy/dx must be undefined,
    or 2y = 0
    y = 0
    then x^2 + 0 = 2 - x
    x^2 + x - 2 = 0
    (x+2)(x-1) = 0
    x = -2 or x = 1

    two vertical tangents, at
    (-2,0) and (1,0)

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  2. Ahem. I think there's an unfortunate typo in there.

    r = 2-cosθ
    r^2 = 2r-rcosθ
    x^2+y^2 = 2√(x^2+y^2) - x

    This complicates things a bit, as there are more vertical tangents. The curve is, after all, (almost) a limaçon, not a circle.

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