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
Aviv S.
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!

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