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Question 1174826: find the equation of parabola which has contact of third order with the conic ax^2 +2hxy + by^2 +2gx +2fy +c =0 at origin.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**1. Understand Contact of Third Order**
* Two curves have contact of third order at a point if they have the same value, the same first derivative, the same second derivative, and the same third derivative at that point.
**2. Parabola Equation**
* Since the parabola has contact at the origin, and we need a general form, we'll use the form:
* y = Ax² + Bx³
* Note that the parabola must pass through the origin (0,0) and the tangent at the origin must be along the x axis.
**3. Conic Equation**
* The given conic is:
* ax² + 2hxy + by² + 2gx + 2fy + c = 0
**4. Contact at the Origin**
* Since both curves pass through the origin (0, 0), we have c = 0.
* Thus the conic equation becomes:
* ax² + 2hxy + by² + 2gx + 2fy = 0
**5. Derivatives**
* **Parabola:**
* y = Ax² + Bx³
* y' = 2Ax + 3Bx²
* y'' = 2A + 6Bx
* y''' = 6B
* **Conic:**
* To find derivatives, we'll use implicit differentiation.
* ax² + 2hxy + by² + 2gx + 2fy = 0
* Differentiate w.r.t x: 2ax + 2hy + 2hx y' + 2by y' + 2g + 2fy' = 0
* At (0, 0): 2g + 2fy' = 0 => y' = -g/f
* If the tangent is along the x axis then y' = 0, thus g = 0.
* Conic equation now is: ax² + 2hxy + by² + 2fy = 0.
* 2ax + 2hy + 2hxy' + 2byy' + 2fy'=0.
* Differentiate again. 2a + 2hy' + 2hy' + 2hxy'' + 2by'y' + 2byy'' + 2fy'' = 0.
* At (0,0) with g=0 and y'=0, 2a + 2fy''=0, thus y'' = -a/f
* Differentiate again. 2hy'' + 2hy'' + 2hy'' + 2hxy''' + 4by'y'' + 2by'y'' + 2byy''' + 2fy''' = 0
* At (0,0) with g=0 and y'=0, 6hy'' + 2fy''' = 0, thus y''' = -3hy''/f = 3ha/f².
**6. Equate Derivatives at (0, 0)**
* **y'(0):**
* Parabola: 0
* Conic: 0 (since g=0)
* **y''(0):**
* Parabola: 2A
* Conic: -a/f
* Therefore, 2A = -a/f => A = -a/(2f)
* **y'''(0):**
* Parabola: 6B
* Conic: 3ha/f²
* Therefore, 6B = 3ha/f² => B = ha/(2f²)
**7. Parabola Equation**
* Substitute A and B into the parabola equation:
* y = (-a/(2f))x² + (ha/(2f²))x³
* y = (-ax²f + hax³)/(2f²)
* 2f²y = -ax²f + hax³
* 2f²y + ax²f - hax³ = 0
**Final Equation**
The equation of the parabola is:
**2f²y + ax²f - hax³ = 0**
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
The solution in the post by @CPhill is incorrect.
It is incorrect TWO TIMES due to two reasons.
(1) The equation of the parabola is written in incorrect form.
(2) in the solution, @CPhill makes an assumption that the tangent is along x-axis.
The problem nowhere says it, so this assumption is WRONG and irrelevant to the problem.
To be correct, this assumption should not be made.
Google AI gives another solution to this problem (quite long) without making wrong assumptions.
See the link
https://www.google.com/search?q=find+the+equation+of+parabola+which+has+contact+of+third+order+with+the+conic+ax%5E2+%2B2hxy+%2B+by%5E2+%2B2gx+%2B2fy+%2Bc+%3D0+at+origin.&rlz=1C1CHBF_enUS1071US1071&oq=find+the+equation+of+parabola+which+has+contact+of+third+order+with+the+conic+ax%5E2+%2B2hxy+%2B+by%5E2+%2B2gx+%2B2fy+%2Bc+%3D0+at+origin.&gs_lcrp=EgZjaHJvbWUyBggAEEUYOdIBCTE5NTFqMGoxNagCCLACAfEFU-UGEVhaVo0&sourceid=chrome&ie=UTF-8
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