Write in standard form

Question

Write in standard form

16x^2 - y^2 - 96x - 10y + 103 = 0

Answer 1

Hi Alyssa,

Assuming you are referring to standard form for a circle, ellipse, or hyperbola, you need to figure out how to factor the equation into the form A(x-B)^2 ± C(y-D)^2 = E

Looking at the x terms, if we factor out a 16 to get rid of the coefficient on x^2, we get 16(x^2 - 6x)...

To complete that polynomial in the parenthesis so that we can factor it into (x-a)^2, we need an additional term, 9 (since -3 goes into -6 twice and multiplies to 9. If that doesn't make sense, please review factoring). We can add this 9 by adding 144 to both sides, since 16 x 9 = 144 (we need to add it inside the quantity being multiplied by 16).

Now we have 16(x^2 - 6x + 9) - y^2 - 10y + 103 = 144

We can now factor the x terms and also move the 103 over:

16(x-3)^2 - y^2 - 10y = 41

We need to make the y^2 term positive to factor it next, so I multiplied everything by -1:

-16(x-3)^2 + y^2 + 10y = -41

Doing the same for the y terms as we did with the x terms, we can introduce a 25 to both sides:

-16(x-3)^2 + (y + 5)^2 = -16

Divide by -16;

(x-3)^2 - 1/16 * (y + 5)^2 = 1

It appears we have a hyperbola. Feel free to look up the formula for guidance on how you would draw it based on this equation.

Cheers!

Jonathan