How do you find the coordinates of the center, foci, the length of the major and minor axis given #x^2+5y^2+4x-70y+209=0#?

Given:

#x^2+5y^2+4x-70y+209=0#

Group, the x terms together, the y terms together, and move the constant term to the right:

#x^2+4x +5y^2-70y= -209#

Please observe the pattern

#(x-h)^2 = x^2-2hx + h^2#.

To make the x terms look like the pattern, we must insert #+h^2# on the left but, to maintain equality, we must, also add #h^2# to the right:

#x^2+4x+ h^2 +5y^2-70y= -209+ h^2#

We can solve for the value of h, if we set the middle term in the pattern equal to the middle term in the equation:

#-2hx = 4x#

#h = -2#

This allows us to substitute #(x - (-2))^2# for the x terms on the left and 4 for #h^2# on the right:

#(x-(-2))^2 +5y^2-70y= -209+ 4#

Please observe the pattern

#(y-k)^2 = y^2-2ky+k^2#

We must multiply both sides by 5 so that the pattern matches the equation:

#5(y-k)^2 = 5y^2-10ky+5k^2#

This means that we must add #5k^2# to both sides of the equation:

#(x-(-2))^2 +5y^2-70y+5k^2= -209+ 4+ 5k^2#

As we did with h, we can use the middle terms to solve for k:

#-10ky = -70y#

#k = 7#

This means that we can substitute 5(y-7)^2 for 5y^2-70y+5k^2 and 245 for #5k^2# on the right:

#(x-(-2))^2 +5(y-7)^2= -209+ 4+ 245#

Simplify the right:

#(x-(-2))^2 +5(y-7)^2= 40#

Divide both sides by 40:

#(x-(-2))^2/40 +(y-7)^2/8= 1#

Write the denominators as squares:

#(x-(-2))^2/(2sqrt10)^2 +(y-7)^2/(2sqrt2)^2= 1#

Here is the corresponding form:

#(x-h)^2/a^2+(y-k)^2/b^2= 1; a > b" [1]"#

This allows us to obtain the desired information by observation.

The center is:

#(h,k) = (-2,5)#

The left focus is:

#(h - sqrt(a^2-b^2), k) = (-2-sqrt(40+8), 7) #

#(h - sqrt(a^2-b^2), k) = (-2-4sqrt3, 7) #

The right focus is:

#(h + sqrt(a^2-b^2), k) = (-2+4sqrt3, 7) #

The length of the major axis is:

#2a = 4sqrt10#

The length of the minor axis is:

#2b = 4sqrt2#