9x² + 8ly² - 729 = 0
Standarn form:
a=
b=
c=
Orientation=
Center=
Vertices=
Co-Vertices=
Foci=

12lx² + 16y² - 1936 = 0
Standarn form:
a=
b=
c=
Orientation=
Center=
Vertices=
Co-Vertices=
Foci=

Let's solve both of these equations step by step and convert them into the standard form of an ellipse. We will then extract the values for \(a\), \(b\), \(c\), orientation, center, vertices, co-vertices, and foci.

### 1st Equation: \(9x^2 + 8ly^2 - 729 = 0\)

First, rewrite the equation:
\[
9x^2 + 8ly^2 = 729
\]
Now, divide both sides by 729 to normalize the equation:
\[
\frac{x^2}{81} + \frac{y^2}{\frac{729}{8l}} = 1
\]

Thus, we have the standard form of the ellipse:
\[
\frac{x^2}{81} + \frac{y^2}{\frac{729}{8l}} = 1
\]

Now, we can extract the key values from the standard form:

- \(a^2 = 81\), so \(a = 9\)
- \(b^2 = \frac{729}{8l}\), so \(b = \frac{\sqrt{729}}{\sqrt{8l}} = \frac{27}{\sqrt{8l}}\)
- To find \(c\), use the relationship \(c^2 = a^2 - b^2\). Therefore, \(c = \sqrt{a^2 - b^2} = \sqrt{81 - \frac{729}{8l}}\).

#### Summary for \(9x^2 + 8ly^2 - 729 = 0\):
- **a** = 9
- **b** = \( \frac{27}{\sqrt{8l}} \)
- **c** = \( \sqrt{81 - \frac{729}{8l}} \)
- **Orientation** = Since \(a^2 > b^2\), the major axis is horizontal.
- **Center** = (0, 0)
- **Vertices** = \((\pm 9, 0)\)
- **Co-Vertices** = \(\left(0, \pm \frac{27}{\sqrt{8l}}\right)\)
- **Foci** = \((\pm \sqrt{81 - \frac{729}{8l}}, 0)\)

---

### 2nd Equation: \(12lx^2 + 16y^2 - 1936 = 0\)

First, rewrite the equation:
\[
12lx^2 + 16y^2 = 1936
\]
Now, divide both sides by 1936 to normalize the equation:
\[
\frac{x^2}{\frac{1936}{12l}} + \frac{y^2}{121} = 1
\]

Thus, the standard form of the ellipse is:
\[
\frac{x^2}{\frac{1936}{12l}} + \frac{y^2}{121} = 1
\]

Now, we can extract the key values from this equation:

- \(a^2 = 121\), so \(a = 11\)
- \(b^2 = \frac{1936}{12l}\), so \(b = \frac{\sqrt{1936}}{\sqrt{12l}} = \frac{44}{\sqrt{12l}}\)
- To find \(c\), use \(c^2 = a^2 - b^2\). Therefore, \(c = \sqrt{a^2 - b^2} = \sqrt{121 - \frac{1936}{12l}}\).

#### Summary for \(12lx^2 + 16y^2 - 1936 = 0\):
- **a** = 11
- **b** = \( \frac{44}{\sqrt{12l}} \)
- **c** = \( \sqrt{121 - \frac{1936}{12l}} \)
- **Orientation** = Since \(a^2 < b^2\), the major axis is vertical.
- **Center** = (0, 0)
- **Vertices** = \((0, \pm 11)\)
- **Co-Vertices** = \(\left(\pm \frac{44}{\sqrt{12l}}, 0\right)\)
- **Foci** = \((0, \pm \sqrt{121 - \frac{1936}{12l}})\)

---

Let me know if you'd like further details or clarifications!

Here are 5 related questions to expand on this topic:
1. What are the equations of the directrices for each ellipse?
2. How would the eccentricity of these ellipses be calculated?
3. Can you identify the eccentricity range for ellipses and compare it to that of a circle or hyperbola?
4. What would the standard form look like if the ellipses were rotated?
5. How would changing the \(l\) parameter affect the size and shape of these ellipses?

**Tip**: To determine the orientation of an ellipse, compare the denominators in the standard form. The larger denominator indicates the major axis direction!

9x² + 8ly² - 729 = 0 Standarn form: a= b= c= Orientation= Center= Vertices= Co-Vertices= Foci= 12lx² + 16y² - 1936 = 0 Standarn form: a= b= c= Orientation= Center= Vertices= Co-Vertices= Foci=

Explore how to convert the equations 9x² + 8ly² - 729 = 0 and 12lx² + 16y² - 1936 = 0 into the standard form of an ellipse. This step-by-step solution covers the key concepts of ellipses, including the values of a, b, c, orientation, center, vertices, co-vertices, and foci. Suitable for students in Grades 10-12 studying conic sections and algebra.

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