Hi,

64x^2+16y^2+320x-192y+960=0

 64(x +5/2)^2 - 64(25/4)  +  16(y+6)^2 - 16(36) +960 = 0

 64(x +5/2)^2 +  16(y+6)^2 = 16

  4(x+5/2)^2  + (y+6)^2 =1  Yes, good Work.

    This format gives the standard form

 C(-5/2,-6)   V(-3,-6)   &    V(-2,6)  and   V(-5/2,-7)  &  V(-5/2,-5)

sqrt(1-1/4) = sqrt(3/4) = ± sqrt(3)/2  F(-5/2, -6±sqrt(3)/2)  

 Standard Form of an Equation of an Ellipse is  where Pt(h,k) is the center. (a positioned to correspond with major axis)

 a and b  are the respective vertices distances from center and ± are the foci distances from center: a > b

 

Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
 64x^2+16y^2+320x-192y+960=0

need in ellipse form plus foci, verts, and ecc 

***

64x^2+16y^2+320x-192y+960=0

complete the square

64x^2+320x+16y^2-192y+960=0

64(x^2+5+25/4)+16(y^2-12y+36)=-960+400+576

64(x+5/2)^2+16(y-6)^2=16

(x+5/2)^2/(16/64)+(y-6)^2=1

(x+5/2)^2/(1/4)+(y-6)^2=1

This is an equation of ellipse with vertical major axis.

Its standard form: (x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k)=(x,y) coordinates of center

For given equation:

center:(-5/2,6)

a^2=1

a=1

vertices: (-5/2,6±a)=(-5/2, 6±1)= (-5/2,5) and (-5/2,7)

b^2=1/4

b=√(1/4)=1/2

..

c^2=a^2-b^2=1-1/4=3/4

c=√(3/4)≈.87

foci: (-5/2,6±c)=(-5/2, 6±.87)= (-5/2,5.13) and (-5/2,6.87)

..

eccentricity: c/a=c/1≈.87



 

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