2016-3b

Step 1: Group Homogeneous Terms

• Degree 3 (\(\phi_3\)): \(2x^3 – 5x^2y + 4xy^2 – y^3\)
• Degree 2 (\(\phi_2\)): \(6x^2 – 7xy + y^2\)
• Degree 1 (\(\phi_1\)): \(-x + 5y\)

Step 2: Find the Slopes (\(m\)) Put \(x=1, y=m\) in \(\phi_3\): \[ \phi_3(m) = 2 – 5m + 4m^2 – m^3 = 0 \] Solving for \(m\): \[ -(m-1)^2(m-2) = 0 \] The roots are \(m = 1, 1, 2\).

Step 3: For Non-Repeated Root (\(m = 2\)) Using \(c = -\frac{\phi_2(m)}{\phi’_3(m)}\):

• \(\phi_2(2) = 6 – 7(2) + (2)^2 = -4\)
• \(\phi’_3(m) = -5 + 8m – 3m^2 \implies \phi’_3(2) = -1\)

\[ c = -(-4 / -1) = -4 \] Asymptote 1: \(2x – y – 4 = 0\)

Step 4: For Repeated Roots (\(m = 1, 1\)) Using the quadratic formula for \(c\): \[ \frac{c^2}{2!}\phi”_3(1) + \frac{c}{1!}\phi’_2(1) + \phi_1(1) = 0 \] Values at \(m=1\):

• \(\phi”_3(1) = 8 – 6(1) = 2\)
• \(\phi’_2(1) = -7 + 2(1) = -5\)
• \(\phi_1(1) = -1 + 5(1) = 4\)

Equation: \(c^2 – 5c + 4 = 0 \implies (c-1)(c-4) = 0\)
Asymptotes 2 & 3: \(x – y + 1 = 0\) and \(x – y + 4 = 0\)