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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

— — — — — — — — — — — — —

- ?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 .