using completing the square, write the equation $-4x^{2}+9y^{2}+32x+36y-64=0$ in standard form. show each…

using completing the square, write the equation $-4x^{2}+9y^{2}+32x+36y-64=0$ in standard form. show each step in the process.\n(6 points)

using completing the square, write the equation $-4x^{2}+9y^{2}+32x+36y-64=0$ in standard form. show each step in the process.\n(6 points)

Answer

Explanation:

Step1: Group x, y terms, move constant

$-4x^2 + 32x + 9y^2 + 36y = 64$

Step2: Factor coefficients of squared terms

$-4(x^2 - 8x) + 9(y^2 + 4y) = 64$

Step3: Complete the square for x

$x^2-8x$: add $(\frac{-8}{2})^2=16$; adjust left side: $-4\times16=-64$

Step4: Complete the square for y

$y^2+4y$: add $(\frac{4}{2})^2=4$; adjust left side: $9\times4=36$

Step5: Add adjustments to right side

$-4(x^2-8x+16) + 9(y^2+4y+4) = 64 - 64 + 36$

Step6: Simplify perfect squares and constants

$-4(x-4)^2 + 9(y+2)^2 = 36$

Step7: Divide by 36 to get 1 on right

$\frac{9(y+2)^2}{36} - \frac{4(x-4)^2}{36} = \frac{36}{36}$ $\frac{(y+2)^2}{4} - \frac{(x-4)^2}{9} = 1$

Answer:

$\frac{(y+2)^2}{4} - \frac{(x-4)^2}{9} = 1$