Question

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.

Answer 1

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!