<PRE><B>Question 1174826</B>
Let&#39;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&#39;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&#39; = 2Ax + 3Bx²
    * y&#39;&#39; = 2A + 6Bx
    * y&#39;&#39;&#39; = 6B

* **Conic:**
    * To find derivatives, we&#39;ll use implicit differentiation.
    * ax² + 2hxy + by² + 2gx + 2fy = 0
    * Differentiate w.r.t x: 2ax + 2hy + 2hx y&#39; + 2by y&#39; + 2g + 2fy&#39; = 0
    * At (0, 0): 2g + 2fy&#39; = 0 =&gt; y&#39; = -g/f
    * If the tangent is along the x axis then y&#39; = 0, thus g = 0.
    * Conic equation now is: ax² + 2hxy + by² + 2fy = 0.
    * 2ax + 2hy + 2hxy&#39; + 2byy&#39; + 2fy&#39;=0.
    * Differentiate again. 2a + 2hy&#39; + 2hy&#39; + 2hxy&#39;&#39; + 2by&#39;y&#39; + 2byy&#39;&#39; + 2fy&#39;&#39; = 0.
    * At (0,0) with g=0 and y&#39;=0, 2a + 2fy&#39;&#39;=0, thus y&#39;&#39; = -a/f
    * Differentiate again. 2hy&#39;&#39; + 2hy&#39;&#39; + 2hy&#39;&#39; + 2hxy&#39;&#39;&#39; + 4by&#39;y&#39;&#39; + 2by&#39;y&#39;&#39; + 2byy&#39;&#39;&#39; + 2fy&#39;&#39;&#39; = 0
    * At (0,0) with g=0 and y&#39;=0, 6hy&#39;&#39; + 2fy&#39;&#39;&#39; = 0, thus y&#39;&#39;&#39; = -3hy&#39;&#39;/f = 3ha/f².

**6. Equate Derivatives at (0, 0)**

* **y&#39;(0):**
    * Parabola: 0
    * Conic: 0 (since g=0)
* **y&#39;&#39;(0):**
    * Parabola: 2A
    * Conic: -a/f
    * Therefore, 2A = -a/f =&gt; A = -a/(2f)
* **y&#39;&#39;&#39;(0):**
    * Parabola: 6B
    * Conic: 3ha/f²
    * Therefore, 6B = 3ha/f² =&gt; 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**
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