write equation
-4x^2+9y^2+32x+36y-64=0
standard form show each step
sorry -- you went astray at the very first step. It should have been
-4x^2 + 9y^2 + 32x + 36y = 64
-4(x^2 - 8x) + 9(y^2 + 4y) = 64
and yet somehow you still arrived at the correct answer ...
You are correct, I made a mistake in the initial step but still ended up with the correct answer. Apologies for the confusion.
To write the equation -4x^2 + 9y^2 + 32x + 36y - 64 = 0 in standard form, we need to complete the square for both the x and y terms.
Step 1: Rearrange the equation
-4x^2 + 9y^2 + 32x + 36y = 64
Step 2: Group the x terms and complete the square
-4(x^2 - 8x) + 9y^2 + 32x + 36y = 64
Step 3: Add half of the coefficient of x squared term, (-8/2)^2 = 16, inside the parentheses and also add the same value multiplied by -4 outside the parentheses to maintain the equation's balance.
-4(x^2 - 8x + 16) + 9y^2 + 32x + 36y = 64 - 4(16)
-4(x - 4)^2 + 9y^2 + 32x + 36y = 0
Step 4: Group the y terms and complete the square
-4(x - 4)^2 + 9(y^2 + 4y) = 0
Step 5: Add half of the coefficient of y term, (4/2)^2 = 4, inside the parentheses and also add the same value multiplied by 9 outside the parentheses to maintain the equation's balance.
-4(x - 4)^2 + 9(y^2 + 4y + 4) = 9(4)
-4(x - 4)^2 + 9(y + 2)^2 = 36
Step 6: Divide both sides of the equation by 36 to isolate the remaining constant term.
-4(x - 4)^2/36 + 9(y + 2)^2/36 = 36/36
-(x - 4)^2/9 + (y + 2)^2/4 = 1
The equation -4x^2 + 9y^2 + 32x + 36y - 64 = 0 is now in standard form: -(x - 4)^2/9 + (y + 2)^2/4 = 1