What are the solutions to the system of conics?

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  • 4 weeks ago

    From  (1),  y^2 = 2x^2 + 4

    So using also  (2),

    2x^2 + 4 = 8x + 8 ,

    2x^2 - 8x - 4 = 0 ,

    x^2 - 4x        = 2 ,

    x^2 - 4x + 4 = 2 + 4 ,

    (x + 2)^2 = 6 ,

    x = - 2 ± √6 .

    Then

    y^2 = 8 (-1 ± √6)

    \  y = ± 2√2 √(-1 ± √6) , four values   ...   ?

    But  -1 - √6 < 0 , so cannot be a value of  y^2 .

    Thus, the solutions points are

          (-2 + √6 . ± 2√2 √(-1 + √6) ) .

  • y^2 / 4 - x^2 / 2 = 1

    y^2 - 2x^2 = 4

    y^2 = 8 * (x + 1)

    8 * (x + 1) - 2x^2 = 4

    4 * (x + 1) - x^2 = 2

    4x + 4 - x^2 = 2

    -x^2 + 4x + 2 = 0

    x^2 - 4x - 2 = 0

    x^2 - 4x = 2

    x^2 - 4x + 4 = 2 + 4

    (x - 2)^2 = 6

    x - 2 = +/- sqrt(6)

    x = 2 +/- sqrt(6)

    When choosing between which function to substitute x into, always choose the one with the lower degree.  The higher degree formulas will give extraneous answers

    y^2 = 8 * (x + 1)

    y^2 = 8 * (2 +/- sqrt(6) + 1)

    y^2 = 8 * (3 +/- sqrt(6))

    y^2 = 4 * (6 +/- 2 * sqrt(6))

    y = +/- 2 * sqrt(6 +/- 2 * sqrt(6))

    (2 + sqrt(6) , 2 * sqrt(6 + 2 * sqrt(6)))

    (2 + sqrt(6) , -2 * sqrt(6 + 2 * sqrt(6)))

    (2 - sqrt(6) , 2 * sqrt(6 - 2 * sqrt(6)))

    (2 - sqrt(6) , -2 * sqrt(6 - 2 * sqrt(6)))

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