Pope
Lv 7
Pope asked in Science & MathematicsMathematics · 2 years ago

Two parabolas having the same intercepts?

In the x-y plane, two distinct parabolas share these same axis intercepts.

Exactly two x-intercepts: (4, 0), (36,0)

Exactly one y-intercept: (0, 9)

Derive the equations of both parabolas.

Do be careful. Not all parabolas have vertical axes.

Update:

Actually, upon further consideration, there are three parabolas sharing those properties. I will still settle for two though.

2 Answers

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  • ?
    Lv 7
    2 years ago
    Favourite answer

    The more complex two will have to be vertically tangent at (0,9).

    Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = 0

    2Ax + 2Bxy + 2By + 2Cyy + 2D + 2Ey + 0 = 0

    (Bx + Cy + E)y = -(Ax + By + D)

    y = -(Ax + By + D)/(Bx + Cy + E)

    dx/dy = 0 at (0,9) so -(9C + E)/(9B + D) = 0, so E = -9C

    (4, 0): 16A + 8D + F = 0

    (36,0): 1296A + 72D + F = 0 so D = -20 A, F = 144 A

    (0,9): 81C + 18E + F = 0 so 81C - 162C + F = 0, so F = 81C

    B^2=AC and F = 81C and D = -20A and F = 144A and E = -9C

    9x² + 24xy + 16y² - 360x - 288y + 1296 = 0

    9x² - 24xy + 16y² - 360x - 288y + 1296 = 0

    — — — — — — — — — — — — —

    The more obvious parabola is y = (9/(4 × 36))(x - 4)(x - 36)

    which is y = (1/16)(x - 20)² - 16 or y = (1/16)x² - (5/2)x + 9

    — — — — — — — — — — — — —

    Graph: https://www.desmos.com/calculator/qn2jorjvac

  • atsuo
    Lv 6
    2 years ago

    A parabola is a kind of conic section . Its equation is

    Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = 0 ---(#1)

    If the conic section is a parabola , then AC = B^2 .

    You said the requested parabolas have exactly 3 x- or y- common intercepts

    (4,0), (36,0) and (0,9) , so they do not passes through (0,0) .

    If F = 0 then the curve of (#1) passes through (0,0) , so

    the requested parabolas do not pass through (0,0) . That is , F ≠ 0 .

    So we can divide (#1) by F ,

    (A/F)x^2 + 2(B/F)xy + (C/F)y^2 + 2(D/F)x + 2(E/F)y + 1 = 0 ---(#2)

    To simplify , rewrite (#2) as (#3) .

    Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + 1 = 0 ---(#3)

    Substitute y = 0 into (#3) , we find

    Ax^2 + 2Dx + 1 = 0 ---(#4)

    (#4) has two roots x = 4 and x = 36 , so

    [-2D + √(4D^2 - 4A)]/(2A) = [-D + √(D^2 - A)]/A = 36 ---(#5)

    [-2D - √(4D^2 - 4A)]/(2A) = [-D - √(D^2 - A)]/A = 4 ---(#6)

    Add (#5) and (#6) ,

    -2D/A = 40

    D = -20A ---(#7)

    Subtract (#6) from (#5) ,

    2√(D^2 - A)/A = 32

    √(D^2 - A) = 16A

    D^2 - A = 256A^2

    Substitute (#7) into it ,

    400A^2 - A = 256A^2

    144A^2 - A = 0

    A(144A - 1) = 0

    A = 0 or A = 1/144

    But if A = 0 then (#4) does not become a quadratic equation .

    So A must be 1/144 . Substitute it into (#7) ,

    D = -5/36

    Next , substitute x = 0 into (#3) , we find

    Cy^2 + 2Ey + 1 = 0 ---(#8)

    (#8) has only one root y = 9 . So

    81C + 18E + 1 = 0 ---(#9)

    And two cases exist .

    Case1. (#8) is not a quadratic equation

    That is , C = 0 . So (#9) becomes

    18E + 1 = 0

    E = -1/18

    We find one solution (A,B,C,D,E) = (1/144, B, 0, -5/36, -1/18) .

    And it must be AC = B^2 , so B = 0 . Therefore (#3) becomes

    (1/144)x^2 - (5/18)x - (1/9)y + 1 = 0 ---(#10)

    This is the equation of the 1st parabola . It can be simplified to

    y = (1/16)x^2 - (5/2)x + 9 ---(#10A)

    Case2. (#8) is a quadratic equation

    That is , C ≠ 0 and it has one double root . So the discriminant = 0 .

    (2E)^2 - 4*C = 0

    4E^2 = 4C

    E^2 = C ---(#11)

    Substitute (#11) into (#9) ,

    81E^2 + 18E + 1 = 0

    (9E + 1)^2 = 0

    E = -1/9

    Substitute it into (#11) ,

    C = 1/81

    So we find another solution (A,B,C,D,E) = (1/144, B, 1/81, -5/36, -1/9) .

    And it must be AC = B^2 so B = ±1/√(144*81) = ±1/108 .

    Therefore (#3) becomes

    (1/144)x^2 ± (1/54)xy + (2/81)y^2 - (5/18)x - (2/9)y + 1 = 0 ---(#12)

    *** (2/81) is a typo , (1/81) is correct . ***

    This is the equation of the 2nd and 3rd parabolas .

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