Axes of conic 14x²−4xy+11y²−44x−58y+71=0

Here a = 14, h = −2, b = 11, so h² − ab = 4 − 154 < 0: the conic is an ellipse (and Δ ≠ 0, so it is not degenerate).

Centre

14x − 2y − 22 = 0 and −2x + 11y − 29 = 0. Solving: x = 2, y = 3. Centre: (2, 3).

Shift origin

Constant: g·x₀ + f·y₀ + c = (−22)(2) + (−29)(3) + 71 = −60, so

14X² − 4XY + 11Y² = 60

Rotate

The coefficients of the squares after rotation are the roots of λ² − (a+b)λ + (ab − h²) = 0:

λ² − 25λ + 150 = 0, so λ = 10 and 15.

The new equation is 10u² + 15v² = 60, i.e.

u²/6 + v²/4 = 1

where u is along the eigenvector of λ = 10, that is the direction (1, 2)/√5, and v along (−2, 1)/√5.

Results

• Semi-axes: a = √6 (major), b = 2 (minor). Lengths of axes: 2√6 and 4.
• Eccentricity: e² = 1 − b²/a² = 1 − 4/6 = 1/3, so e = 1/√3.
• Latus rectum: 2b²/a = 8/√6 = 4√6/3.
• Foci: ae = √(a² − b²) = √2 from the centre along (1, 2)/√5, giving (2 ± √10/5, 3 ± 2√10/5).
• Major axis line: 2x − y − 1 = 0; minor axis line: x + 2y − 8 = 0.