How do we reduce 2x^2 + 3y^2 − 4x − 12y + 13 = 0 to standard form?

Complete the squares in x and y, then shift the origin to the centre

Differentiate and set equal to zero only

In X^2/(1/2) + Y^2/(1/3) = 1 which is the major axis and its length?

a^2 = 1/2 is larger, so major axis = sqrt2 along the x-direction

b^2 = 1/3 gives major axis 2/sqrt3

Short answer

Complete squares: 2(x-1)²+3(y-2)²=1. Centre (1,2); major axis length √2, minor 2/√3; eccentricity 1/√3; directrices x=1±√6/2.

Find the centre, length of axes, eccentricity and directrices of the conic (ellipse) 2x² + 3y² − 4x − 12y + 13 = 0.

Complete the squares to bring the conic to the standard form (x−h)²/a² + (y−k)²/b² = 1.

2x² − 4x + 3y² − 12y + 13 = 0

2(x² − 2x) + 3(y² − 4y) = −13

2(x−1)² + 3(y−2)² = −13 + 2 + 12 = 1

Shift the origin to (1, 2): X = x − 1, Y = y − 2. Then

2X² + 3Y² = 1, that is X²/(1/2) + Y²/(1/3) = 1

So a² = 1/2 (major, since 1/2 > 1/3) and b² = 1/3.

Centre: (1, 2).
Axes: major axis 2a = 2/√2 = √2 (parallel to the x-axis); minor axis 2b = 2/√3.
Eccentricity: e² = 1 − b²/a² = 1 − (1/3)/(1/2) = 1/3, so e = 1/√3.
Foci: ae = √(1/2) · (1/√3) = 1/√6, so the foci are (1 ± 1/√6, 2).
Directrices: X = ±a/e = ±(1/√2)·√3 = ±√(3/2), that is x − 1 = ±√6/2, so

x = 1 ± √6/2

Check: a² − b² = 1/2 − 1/3 = 1/6 = (ae)², which matches ae = 1/√6.

✓ Answer checked by Claude Opus 5.5 (high effort)