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Let's solve this problem step-by-step.\r
\n" ); document.write( "\n" ); document.write( "**1. Understand Contact of Third Order**\r
\n" ); document.write( "\n" ); document.write( "* 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.\r
\n" ); document.write( "\n" ); document.write( "**2. Parabola Equation**\r
\n" ); document.write( "\n" ); document.write( "* Since the parabola has contact at the origin, and we need a general form, we'll use the form:
\n" ); document.write( " * y = Ax² + Bx³\r
\n" ); document.write( "\n" ); document.write( " * Note that the parabola must pass through the origin (0,0) and the tangent at the origin must be along the x axis.\r
\n" ); document.write( "\n" ); document.write( "**3. Conic Equation**\r
\n" ); document.write( "\n" ); document.write( "* The given conic is:
\n" ); document.write( " * ax² + 2hxy + by² + 2gx + 2fy + c = 0\r
\n" ); document.write( "\n" ); document.write( "**4. Contact at the Origin**\r
\n" ); document.write( "\n" ); document.write( "* Since both curves pass through the origin (0, 0), we have c = 0.
\n" ); document.write( "* Thus the conic equation becomes:
\n" ); document.write( " * ax² + 2hxy + by² + 2gx + 2fy = 0\r
\n" ); document.write( "\n" ); document.write( "**5. Derivatives**\r
\n" ); document.write( "\n" ); document.write( "* **Parabola:**
\n" ); document.write( " * y = Ax² + Bx³
\n" ); document.write( " * y' = 2Ax + 3Bx²
\n" ); document.write( " * y'' = 2A + 6Bx
\n" ); document.write( " * y''' = 6B\r
\n" ); document.write( "\n" ); document.write( "* **Conic:**
\n" ); document.write( " * To find derivatives, we'll use implicit differentiation.
\n" ); document.write( " * ax² + 2hxy + by² + 2gx + 2fy = 0
\n" ); document.write( " * Differentiate w.r.t x: 2ax + 2hy + 2hx y' + 2by y' + 2g + 2fy' = 0
\n" ); document.write( " * At (0, 0): 2g + 2fy' = 0 => y' = -g/f
\n" ); document.write( " * If the tangent is along the x axis then y' = 0, thus g = 0.
\n" ); document.write( " * Conic equation now is: ax² + 2hxy + by² + 2fy = 0.
\n" ); document.write( " * 2ax + 2hy + 2hxy' + 2byy' + 2fy'=0.
\n" ); document.write( " * Differentiate again. 2a + 2hy' + 2hy' + 2hxy'' + 2by'y' + 2byy'' + 2fy'' = 0.
\n" ); document.write( " * At (0,0) with g=0 and y'=0, 2a + 2fy''=0, thus y'' = -a/f
\n" ); document.write( " * Differentiate again. 2hy'' + 2hy'' + 2hy'' + 2hxy''' + 4by'y'' + 2by'y'' + 2byy''' + 2fy''' = 0
\n" ); document.write( " * At (0,0) with g=0 and y'=0, 6hy'' + 2fy''' = 0, thus y''' = -3hy''/f = 3ha/f².\r
\n" ); document.write( "\n" ); document.write( "**6. Equate Derivatives at (0, 0)**\r
\n" ); document.write( "\n" ); document.write( "* **y'(0):**
\n" ); document.write( " * Parabola: 0
\n" ); document.write( " * Conic: 0 (since g=0)
\n" ); document.write( "* **y''(0):**
\n" ); document.write( " * Parabola: 2A
\n" ); document.write( " * Conic: -a/f
\n" ); document.write( " * Therefore, 2A = -a/f => A = -a/(2f)
\n" ); document.write( "* **y'''(0):**
\n" ); document.write( " * Parabola: 6B
\n" ); document.write( " * Conic: 3ha/f²
\n" ); document.write( " * Therefore, 6B = 3ha/f² => B = ha/(2f²)\r
\n" ); document.write( "\n" ); document.write( "**7. Parabola Equation**\r
\n" ); document.write( "\n" ); document.write( "* Substitute A and B into the parabola equation:
\n" ); document.write( " * y = (-a/(2f))x² + (ha/(2f²))x³
\n" ); document.write( " * y = (-ax²f + hax³)/(2f²)
\n" ); document.write( " * 2f²y = -ax²f + hax³
\n" ); document.write( " * 2f²y + ax²f - hax³ = 0\r
\n" ); document.write( "\n" ); document.write( "**Final Equation**\r
\n" ); document.write( "\n" ); document.write( "The equation of the parabola is:\r
\n" ); document.write( "\n" ); document.write( "**2f²y + ax²f - hax³ = 0**
\n" ); document.write( " \n" ); document.write( "