Find the standard equation of ellipse with center at (h,k).9×²+16y²-12×-64y=71

To find the standard equation of the ellipse, we need to rewrite the given equation in standard form.

The given equation is:
\[
9x^2 + 16y^2 - 12x - 64y = 71
\]

### Step 1: Rearrange the terms
Group the \(x\) and \(y\) terms together:
\[
9x^2 - 12x + 16y^2 - 64y = 71
\]

### Step 2: Complete the square for both \(x\) and \(y\) terms.

#### Completing the square for the \(x\) terms:
The \(x\) terms are \(9x^2 - 12x\).

Factor out the coefficient of \(x^2\) (which is 9):
\[
9(x^2 - \frac{4}{3}x)
\]
Now, complete the square inside the parentheses. Take half of the coefficient of \(x\) (which is \(-\frac{4}{3}\)), square it, and add and subtract it inside the parentheses:
\[
x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2
\]
Simplify:
\[
x^2 - \frac{4}{3}x + \frac{4}{9} - \frac{4}{9}
\]
This can be rewritten as:
\[
9\left[(x - \frac{2}{3})^2 - \frac{4}{9}\right]
\]
Distribute the 9:
\[
9(x - \frac{2}{3})^2 - 4
\]

#### Completing the square for the \(y\) terms:
The \(y\) terms are \(16y^2 - 64y\).

Factor out the coefficient of \(y^2\) (which is 16):
\[
16(y^2 - 4y)
\]
Complete the square inside the parentheses. Take half of the coefficient of \(y\) (which is \(-4\)), square it, and add and subtract it:
\[
y^2 - 4y + 4 - 4
\]
This can be rewritten as:
\[
16\left[(y - 2)^2 - 4\right]
\]
Distribute the 16:
\[
16(y - 2)^2 - 64
\]

### Step 3: Substitute the completed squares back into the equation
\[
9(x - \frac{2}{3})^2 - 4 + 16(y - 2)^2 - 64 = 71
\]
Simplify by moving the constants to the right side:
\[
9(x - \frac{2}{3})^2 + 16(y - 2)^2 = 71 + 4 + 64
\]
\[
9(x - \frac{2}{3})^2 + 16(y - 2)^2 = 139
\]

### Step 4: Divide the equation by 139 to express it in standard form
\[
\frac{9(x - \frac{2}{3})^2}{139} + \frac{16(y - 2)^2}{139} = 1
\]

The equation is now in the standard form of an ellipse:
\[
\frac{(x - \frac{2}{3})^2}{\frac{139}{9}} + \frac{(y - 2)^2}{\frac{139}{16}} = 1
\]

This is the standard form of the ellipse with the center at \(\left(\frac{2}{3}, 2\right)\).

Do you want more details, or have any questions? 

Here are five related questions you might find helpful:
1. How do you determine the major and minor axes of an ellipse?
2. How can you find the foci of an ellipse given its standard form?
3. What changes in the equation of an ellipse when the axes are rotated?
4. Can the standard form of an ellipse be derived from its general form?
5. How does the equation change if the ellipse is centered at the origin?

**Tip:** Completing the square is a powerful technique not just for ellipses, but for solving quadratic equations and analyzing conic sections in general.

Find the standard equation of ellipse with center at (h,k). 9x² + 16y² - 12x - 64y = 71

Learn how to find the standard equation of an ellipse with center at (h, k) by completing the square. This step-by-step solution simplifies 9x² + 16y² - 12x - 64y = 71 into standard form. Ideal for students in Grades 9-12 learning about conic sections and ellipses.

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