Totally confused-learned this today and I don't get it-Please help
Write the standard form of 9x^2 + 4y^2-72x+16y+124=0
I think the first step is to take the constant to right side so I have:
9x^2 + 4y^2-72x+16y=124
But then I'm lost!
To write the given equation in standard form, we need to complete the square for both the x and y variables. Here's how you can go about it:
1. Start by grouping the x and y terms together:
(9x^2 - 72x) + (4y^2 + 16y) = 124
2. Next, factor out the coefficient of x^2 and y^2 from their respective groups:
9(x^2 - 8x) + 4(y^2 + 4y) = 124
3. Now, we need to complete the square for both the x and y terms individually. Let's begin with the x terms. Take half of the coefficient of x (-8) and square it (-8/2)^2 = 16:
9(x^2 - 8x + 16) + 4(y^2 + 4y) = 124 + 9(16)
4. We also need to add the same value inside the parentheses for the y terms. Take half of the coefficient of y (4) and square it (4/2)^2 = 4:
9(x^2 - 8x + 16) + 4(y^2 + 4y + 4) = 124 + 9(16) + 4(4)
5. Simplify the equation using the distributive property:
9(x^2 - 8x + 16) + 4(y^2 + 4y + 4) = 124 + 144 + 16
6. Simplify further:
9(x - 4)^2 + 4(y + 2)^2 = 284
Now, the equation is in standard form, which is:
A(x - h)^2 + B(y - k)^2 = C
In this case, A = 9, B = 4, h = 4, k = -2, and C = 284.