A curve is represented by $C = 21x^2 - 6xy + 29y^2 + 6x - 58y - 151 = 0$. Eccentricity of curve is:
Step-by-Step Solution
Key Concept: Rotated conics require completing the square to identify the principal axes and convert to standard ellipse form.
The equation $21x^2-6xy+29y^2+6x-58y-151=0$ is rewritten by completing the square as $3(x-3y+3)^2+(3x+y-1)^2=180$. This transforms to standard form $\frac{(x-3y+3)^2}{60}+\frac{(3x+y-1)^2}{90}=1$, representing an ellipse with semi-axes lengths $6$ and $2\sqrt{6}$. The eccentricity is $e=\frac{1}{\sqrt{3}}$, and the major and minor axes lie on $x-3y+3=0$ and $3x+y-1=0$ respectively, intersecting at centre $(0,1)$.
Correct Answer: 2