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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.
Actually, upon further consideration, there are three parabolas sharing those properties. I will still settle for two though.
2 Answers
- Φ² = Φ+1Lv 72 years agoFavorite 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
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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
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- ?Lv 62 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 .