Question

1. find the standard form equation for the ellipse.

$25x^2 + 16y^2 + 150x - 64y = 111$
$\frac{(x+3)^2}{5^2}+\frac{(y-2)^2}{4^2}=1$
$\frac{(x-3)^2}{4^2}+\frac{(y+2)^2}{5^2}=1$
$\frac{(x+3)^2}{4^2}+\frac{(y-2)^2}{5^2}=1$
$\frac{(x+3)^2}{4^2}-\frac{(y+2)^2}{5^2}=1$

Explanation:

Step1: Rearrange the given equation

$25x^2 + 150x + 16y^2 - 64y = 111$

Step2: Group x and y terms

$25(x^2 + 6x) + 16(y^2 - 4y) = 111$

Step3: Complete the square for x

$x^2+6x=(x+3)^2-9$, so $25[(x+3)^2-9] = 25(x+3)^2 - 225$

Step4: Complete the square for y

$y^2-4y=(y-2)^2-4$, so $16[(y-2)^2-4] = 16(y-2)^2 - 64$

Step5: Substitute back and simplify

$25(x+3)^2 - 225 + 16(y-2)^2 - 64 = 111$
$25(x+3)^2 + 16(y-2)^2 = 111 + 225 + 64$
$25(x+3)^2 + 16(y-2)^2 = 400$

Step6: Divide by 400 to get standard form

$\frac{25(x+3)^2}{400} + \frac{16(y-2)^2}{400} = 1$
$\frac{(x+3)^2}{5^2} + \frac{(y-2)^2}{4^2} = 1$

Answer:

$\boldsymbol{\frac{(x+3)^2}{5^2} + \frac{(y-2)^2}{4^2} = 1}$ (corresponding to the first option)