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Lv 7
? asked in Science & MathematicsMathematics · 1 year ago

What point on the curve y^2 = x + 4 is closest to (4, -2)?

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    x = y^2 - 4

    y = y

    d^2 = (y - (-2))^2 + (x - 4)^2

    d^2 = (y + 2)^2 + (y^2 - 4 - 4)^2

    d^2 = y^2 + 4y + 4 + (y^2 - 8)^2

    d^2 = y^2 + 4y + 4 + y^4 - 16y^2 + 64

    d^2 = y^4 - 15y^2 + 4y + 68

    Now use a derivative

    2d * dd/dy = 4y^3 - 30y + 4

    dd/dy = 0

    2d * 0 = 4y^3 - 30y + 4

    0 = 4y^3 - 30y + 4

    0 = 2y^3 - 15y + 2

    https://www.wolframalpha.com/input/?i=0+%3D+2y%5E3...

    y = -2.803 , 0.13365 , 2.6693

    d^2 = y^4 - 15y^2 + 4y + 68

    y = -2.803 , d = ‭0.8156670046538599565396096108566‬

    y = 0.13365 , d = ‭8.2623836890491289993182248560078‬

    y = 2.6693 , d = ‭4.750547669364343943063299623439‬

    x = ‭3.12516249‬ , y = 2.6693 , d = 4.75055

    (3.13 , 2.67) with a distance of 4.75

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