How do you change x^2 + 4y^2 - 4x + 8y - 60 = 0x2+4y2−4x+8y−60=0 into standard form?

1 Answer
Alan P.
Mar 20, 2016

((x-2)^2)/((2sqrt(17))^2)+((y+1)^2)/((sqrt(17)/2)^2)=1(x−2)2(2√17)2+(y+1)2(√172)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)=1XXX(x−h)2a2+(y−k)2b2=1

Given:
color(white)("XXX")x^2+4y^2-4x+8y-60=0XXXx2+4y2−4x+8y−60=0

Re-arranging the terms to group the xx terms, the yy terms, and shift the constant to the right side:
color(white)("XXX")x^2-4x+4y^2+8y=60XXXx2−4x+4y2+8y=60

Extract the common factor from the yy terms and complete the squares
color(white)("XXX")x^2-4x+4+4(y^2+2y+1)=60+4+4XXXx2−4x+4+4(y2+2y+1)=60+4+4

Re-writing as squared binomials and simplifying
color(white)("XXX")(x-2)^2+4(y+1)^2=68XXX(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)=1XXX(x−2)2(2√17)2+(y+1)2(√172)2=1

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