Question

Find the eccentricity of 8x^2 + 6y^2 - 32x + 24y + 8 = 0

Answer 1

1) 
x 2 + y 2 - 10x - 8y + 1 = 0. 
(x^2-10x+25-25) +(y^2-8y+16-16) + 1 = 0 
(x-5)^2 -25 + (y-4)^2 -16 + 1 = 0 
(x-5)^2+(y-4)^2 = 40 

center (5,4); radius = sqrt(40) 

2) 
8x^2+ 6y 2 - 32x + 24y + 8 = 0. 
8(x^2-4x+4-4) +6(y^2+4y+4-4) +8=0 
8(x-2)^2 -32 + 6(y+2)^2 -24 + 8 = 0 
8(x-2)^2+6(y+2)^2 = 48 
divide throughout by 48 
(x-2)^2 /6 + (y+2)^2 /8 = 1 

Ellipse with center (2,-2) 
a=sqrt(6) 
b=sqrt(8) 
c^2= 8-6 = 2 
c= sqrt(2) 
eccentricity = c/a = sqrt(2)/sqrt(6) 

3) 
y=x^2-12x+36-36 
y=(x-6)^2 - 36 
Vertex is (6,-36) 

(x-h)^2=4p(y-k) 
(x-6)^2 = 4p(y+36) 
4p=1 
p=1/4 
focus : (h, k+p) = (6, -36+1/4) = (6, -143/4) 

4) 
focus lies on a vertical line, so the major axis is parallel to the y-axis 
(x-h)^2/a^2 +(y-b)^2/b^2 = 1 
h=-2 
k=0 
(x+2)^2/a^2+y^2/b^2 = 1 
2b=20 
b=10 
b^2=100 
(x+2)^2/a^2 +y^2/100 = 1 


e=c/a 
c/a = 4/5 
c=(4/5) a 
c^2 = 16/25 a^2 
c^2 = b^2-a^2 
(16/25) a^2 = 100 - a^2 
a^2(16/25+1) = 100 
41a^2/25 = 100 
a^2=2500/41 
a= sqrt(2500/41) 

(x+2)^2/a^2 +y^2/100 = 1 
(x+2)^2 /[2500/41] + y^2/100 = 1