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The conic 6x^2 - 4xy + 9y^2 - 20x + 20y - 20 = 0 is an ellipse. If…

Question

The conic 6x^2 - 4xy + 9y^2 - 20x + 20y - 20 = 0 is an ellipse. If the square of its eccentricity equals m/n in lowest terms, find m+n.

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Step-by-step solution

Step 1 — translate to the centre. Set the partial derivatives to zero:

F_x = 12x - 4y - 20 = 0, F_y = -4x + 18y + 20 = 0.
Solving: from the first 3x - y = 5; from the second -2x + 9y = -10.
Then y = 3x - 5, substitute: -2x + 9(3x-5) = -10 => 25x = 35 => x = 7/5, y = 3(7/5)-5 = -4/5.
Centre = (7/5, -4/5).

The constant after translation is F(centre).

Evaluating F at the centre gives -42, so in centred coordinates X^T Q X = 42, where Q = [[6,-2],[-2,9]] is the matrix of the quadratic part (off-diagonal entries are half the xy-coefficient).

Step 2 — diagonalise Q. Eigenvalues solve (6-λ)(9-λ) - 4 = 0 => λ^2 - 15λ + 50 = 0 => λ = 5 or λ = 10.
The ellipse becomes 5 X'^2 + 10 Y'^2 = 42, i.e. X'^2/(42/5) + Y'^2/(42/10) = 1.
So the squared semi-axes are 42/5 (major) and 42/10 = 21/5 (minor): a^2 = 42/5, b^2 = 21/5.
Step 3 — eccentricity. e^2 = 1 - b^2/a^2 = 1 - (21/5)/(42/5) = 1 - 1/2 = 1/2.
m/n = 1/2, m+n = 3.

Final answer3

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