https://file.aiquickdraw.com/mathbot/file/1728828275740a6rbw3xk.jpg

Let's break down the given equation and solve the problem step by step.

### Given equation:
\[
16x^2 + 4y^2 - 32 = 0
\]

### Step 1: Convert to standard form
First, add 32 to both sides:
\[
16x^2 + 4y^2 = 32
\]

Now, divide the entire equation by 32 to make it easier to identify the form of the ellipse:
\[
\frac{16x^2}{32} + \frac{4y^2}{32} = 1
\]
\[
\frac{x^2}{2} + \frac{y^2}{8} = 1
\]

This is the standard form of the equation of an ellipse:
\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\]
where \(a^2 = 2\) and \(b^2 = 8\), meaning \(a = \sqrt{2}\) and \(b = 2\sqrt{2}\).

### Step 2: Identify key features of the ellipse

1. **Center:** Since there are no terms involving \(x\) or \(y\) shifts, the center of the ellipse is at \( (0, 0) \).

2. **Vertices:**
   - The vertices are located along the axis corresponding to the larger value of \(b^2\), which is along the \(y\)-axis.
   - The length of the semi-major axis (along \(y\)) is \(b = 2\sqrt{2} \approx 2.83\).
   - Therefore, the vertices are at \( (0, \pm 2\sqrt{2}) \) or approximately \( (0, \pm 2.83) \).

3. **Co-vertices:**
   - The co-vertices are along the \(x\)-axis. The length of the semi-minor axis is \(a = \sqrt{2} \approx 1.41\).
   - Thus, the co-vertices are at \( (\pm \sqrt{2}, 0) \) or approximately \( (\pm 1.41, 0) \).

4. **Foci:**
   - To find the foci, we need to calculate the value of \(c\), where \(c^2 = b^2 - a^2\).
   \[
   c^2 = 8 - 2 = 6 \quad \text{so} \quad c = \sqrt{6} \approx 2.45
   \]
   - The foci are located along the \(y\)-axis at \( (0, \pm \sqrt{6}) \) or approximately \( (0, \pm 2.45) \).

5. **Endpoints of the Latus Rectum:**
   - The endpoints of the latus rectum are located at a distance of \( \frac{b^2}{a} \) from the foci along the \(x\)-axis. 
   \[
   \frac{8}{\sqrt{2}} = 4\sqrt{2} \approx 5.66
   \]
   - The endpoints of the latus rectum are at \( (\pm 5.66, \pm 2.45) \).

### Step 3: Graph of the Ellipse
You can sketch the ellipse with the following points:
- Center at \( (0, 0) \)
- Vertices at \( (0, \pm 2.83) \)
- Co-vertices at \( (\pm 1.41, 0) \)
- Foci at \( (0, \pm 2.45) \)

Would you like further details on any specific part of the solution, or help with visualizing the graph?

Here are some related questions to explore:
1. How is the general form of an ellipse derived?
2. What is the significance of the semi-major and semi-minor axes?
3. How do you calculate the eccentricity of an ellipse?
4. What is the role of the latus rectum in conic sections?
5. How can you generalize the approach to find the equation of an ellipse with a shifted center?

**Tip:** Always check whether the larger denominator is under \(x^2\) or \(y^2\) to determine the orientation of the ellipse.

Change it to standard form and find the value of co-vertices, center, vertices, foci, and endpoints of LR. Draw the graph of the given ellipse equation: 16x^2 + 4y^2 - 32 = 0.

Learn how to convert the ellipse equation 16x^2 + 4y^2 - 32 = 0 into standard form and find key features such as the center, vertices, co-vertices, foci, and endpoints of the latus rectum. Includes a step-by-step breakdown and graphing instructions. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation