Find the standard form of the ellipse given below in general form. 9 x 2 − 72 x + y 2 + 10 y + 160

To convert the given general form of the equation into the standard form of an ellipse, we will complete the square for both $x$- and $y$-terms. Let's solve step by step.

General Form:

$9x^2 - 72x + y^2 + 10y + 160 = 0$

---

Step 1: Group $x$- and $y$-terms, factor constants where necessary.

$9(x^2 - 8x) + (y^2 + 10y) + 160 = 0$

---

Step 2: Complete the square for $x$- and $y$-terms.

For $x$-terms:

$x^2 - 8x$ Take half the coefficient of $x$ (which is $-8$), square it, and add it: $\left(\frac{-8}{2}\right)^2 = 16$ Add and subtract $16$ within the parentheses: $9(x^2 - 8x + 16 - 16) = 9((x - 4)^2 - 16)$

Simplify: $9(x - 4)^2 - 144$

For $y$-terms:

$y^2 + 10y$ Take half the coefficient of $y$ (which is $10$), square it, and add it: $\left(\frac{10}{2}\right)^2 = 25$ Add and subtract $25$ within the parentheses: $(y^2 + 10y + 25 - 25) = (y + 5)^2 - 25$

---

Step 3: Substitute back into the equation.

The equation becomes: $9((x - 4)^2 - 16) + ((y + 5)^2 - 25) + 160 = 0$

Expand: $9(x - 4)^2 - 144 + (y + 5)^2 - 25 + 160 = 0$

Combine constants: $9(x - 4)^2 + (y + 5)^2 - 9 = 0$

---

Step 4: Rearrange into standard form.

Add $9$ to both sides: $9(x - 4)^2 + (y + 5)^2 = 9$

Divide through by $9$ to normalize: $\frac{(x - 4)^2}{1} + \frac{(y + 5)^2}{9} = 1$

---

Final Standard Form:

$\frac{(x - 4)^2}{1} + \frac{(y + 5)^2}{9} = 1$

This represents a vertical ellipse centered at $(4, -5)$ with a semi-major axis length of $3$ and a semi-minor axis length of $1$.

---

Questions to Expand Your Understanding:

1. How do we determine whether the ellipse is vertical or horizontal?
2. What are the lengths of the semi-major and semi-minor axes of this ellipse?
3. How would this process differ if we started with a hyperbola equation?
4. What are the key differences between completing the square for circles and ellipses?
5. Can you visualize the graph of this ellipse? What are its focal points?

Tip: When completing the square, always factor out the leading coefficient of the quadratic term before beginning the process.