A) a parabola
B) a pair of straight lines
C) an ellipse with eccentricity ½
D) an ellipse with eccentricity\[\sqrt{3}/2\]
E) a hyperbola with eccentricity 3/2
Correct Answer: D
Solution :
The given equation is \[4{{x}^{2}}-24x+36+16{{y}^{2}}-32y-12-36\] \[+16-16=0\] \[\Rightarrow \] \[{{(2x-6)}^{2}}+{{(4y-4)}^{2}}=64\] \[\Rightarrow \] \[\frac{{{(x-3)}^{2}}}{16}+\frac{{{(y-1)}^{2}}}{4}=1\] This represents an ellipse and \[{{a}^{2}}=16,\text{ }{{b}^{2}}=4\] \[\therefore \] \[e=\sqrt{1-\frac{4}{16}}=\frac{\sqrt{3}}{2}\]
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