How do you change 16x^2+25y^2-32x+50y+16=0 to standard form?

Standard form as in x^2/a^2+y^2/b^2=1.

I can get as far as completing the square [which gets me 16(x-1)^2+25(y+1)^2=25], but I don't know what to do now since 16 is not divisible by 25. Correct me if I'm wrong, and thanks in advance for the help.

1 Answer
Apr 25, 2018

The equation in standard form is (x-1)^2/(5/4)^2+(y+1)^2/1^2=1.

Explanation:

Your completing the square was correct, and all you have to do is rewrite the fraction as the reciprocal of the reciprocal of itself. For instance, 3x can be rewritten as x/(1/3).

Here's the given equation:

16x^2+25y^2-32x+50y+16=0

16x^2-32x+16+25y^2+50y=0

16(x^2-2x+1)+25y^2+50y=0

16(x^2-2x+1)+25(y^2+2y)=0

16(x^2-2x+1)+25(y^2+2y)+25=25

16(x^2-2x+1)+25(y^2+2y+1)=25

16(x-1)^2+25(y+1)^2=25

Use that trick I mentioned earlier:

(x-1)^2/(1/16)+25(y+1)^2=25

((x-1)^2/(1/16))/25+(y+1)^2/1=1

(x-1)^2/(25/16)+(y+1)^2/1=1

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

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

This means that your a value is 5/4, and the b value is 1. Hope this helped!