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Question: Find the eccentricity of the conic $x^2 - 2xy + y^2 - 8x - 4y + 16 = 0$...

Find the eccentricity of the conic x2−2xy+y2−8x−4y+16=0x^2 - 2xy + y^2 - 8x - 4y + 16 = 0

A

0

B

1

C

2

D

infinity

Answer

1

Explanation

Solution

The given equation of the conic is x2−2xy+y2−8x−4y+16=0x^2 - 2xy + y^2 - 8x - 4y + 16 = 0. This is a general second-degree equation of the form Ax2+2Hxy+By2+2Gx+2Fy+C=0Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0.

Comparing the given equation with the general form, we have: A=1A = 1 2H=−2  ⟹  H=−12H = -2 \implies H = -1 B=1B = 1 2G=−8  ⟹  G=−42G = -8 \implies G = -4 2F=−4  ⟹  F=−22F = -4 \implies F = -2 C=16C = 16

To determine the type of conic, we examine the discriminant H2−ABH^2 - AB. H2−AB=(−1)2−(1)(1)=1−1=0H^2 - AB = (-1)^2 - (1)(1) = 1 - 1 = 0.

When H2−AB=0H^2 - AB = 0, the conic section is a parabola, provided it is not a degenerate case. The condition for degeneracy is Δ=ABC+2FGH−AF2−BG2−CH2≠0\Delta = ABC + 2FGH - AF^2 - BG^2 - CH^2 \neq 0. Let's calculate Δ\Delta: Δ=(1)(1)(16)+2(−2)(−4)(−1)−(1)(−2)2−(1)(−4)2−(16)(−1)2\Delta = (1)(1)(16) + 2(-2)(-4)(-1) - (1)(-2)^2 - (1)(-4)^2 - (16)(-1)^2 Δ=16+(−16)−4−16−16\Delta = 16 + (-16) - 4 - 16 - 16 Δ=16−16−4−16−16=−40\Delta = 16 - 16 - 4 - 16 - 16 = -40 Since Δ=−40≠0\Delta = -40 \neq 0, the equation represents a non-degenerate conic.

As H2−AB=0H^2 - AB = 0, the conic section is a parabola.

The eccentricity (ee) of a parabola is always 1, by definition.

Therefore, the eccentricity of the given conic is 1.