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Question

Find The eccentricity, latus rectum and focus of the following ellipse:
3x2+4y212x8y+4=0

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Solution

Equation of parabola,
3x2+4y212x8y+4=0
(3x212x)+(4y28y)+4=0
3(x24x)+4(y22y)+4=0
3(x24x+44)+4(y22y+11)+4=0
3(x2)212+4(y1)24+4=0
3(x2)2+4(y1)2=12
3(x2)212+4(y1)212=1
(x2)24+(y1)23=1
Comparing with X2a2+Y2b2=1
X=x2,Y=y1,a=2,b=3,here a>b
Eccentricity : b2=a2(1e2)
3=4(1e2)
1e2=34
e2=134=14
e=12
Latus rectum: 2b2a=2×32=3
Coordinated of focus coordinates of focus of eliipse will be (±ae,0)
X=±ae
x2=±2×12=±1
x=2±1
x=3,1
y=0
y1=0y=1
Thus coordinate of focus are (3,1) and (1,1)

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