How do you change #x^2 + 4y^2 - 4x + 8y - 60 = 0# into standard form?

1 Answer
Mar 20, 2016

Answer:

#((x-2)^2)/((2sqrt(17))^2)+((y+1)^2)/((sqrt(17)/2)^2)=1#

Explanation:

The given equation is already in standard polynomial form, so I assume that what is required is to express this in standard elliptical form.
#color(white)("XXX")((x-h)^2)/(a^2) +((y-k)^2)/(b^2)=1#

Given:
#color(white)("XXX")x^2+4y^2-4x+8y-60=0#

Re-arranging the terms to group the #x# terms, the #y# terms, and shift the constant to the right side:
#color(white)("XXX")x^2-4x+4y^2+8y=60#

Extract the common factor from the #y# terms and complete the squares
#color(white)("XXX")x^2-4x+4+4(y^2+2y+1)=60+4+4#

Re-writing as squared binomials and simplifying
#color(white)("XXX")(x-2)^2+4(y+1)^2=68#

Divide through by 68and take roots of denominators
#color(white)("XXX")((x-2)^2)/((2sqrt(17))^2)+((y+1)^2)/((sqrt(17)/2)^2)=1#