Find the center, vertices, foci, and eccentricity of the ellipse. $$ 16 x^{2}+24 y^{2}-16 x+48 y-68=0 $$ center $$ (x, y)=( $$ $$ \square $$ ) vertices $$ \quad(x, y)=( $$ $$ \square $$ )(smaller $$ x $$-value) $$ (x, y)=(\square) \text { (larger } x \text {-value) } $$ foci $$ (x, y)= $$ $$ \square $$ ) (smaller $$ x $$-value) $$ \begin{array}{l} (x, y)=(\square \ \square \end{array} $$ eccentricity

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To find the center, vertices, foci, and eccentricity of the given ellipse, we start by rewriting the equation in standard form. The given equation is:

$$
16 x^{2}+24 y^{2}-16 x+48 y-68=0
$$

### Step 1: Rearranging the Equation

First, we will rearrange the equation to group the $$x$$ and $$y$$ terms:

$$
16x^2 - 16x + 24y^2 + 48y - 68 = 0
$$

### Step 2: Completing the Square

Next, we will complete the square for both $$x$$ and $$y$$.

#### For $$x$$:

1. Factor out the coefficient of $$x^2$$:
   $$
   16(x^2 - x)
   $$
2. Complete the square:
   $$
   x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4}
   $$
   Thus,
   $$
   16\left((x - \frac{1}{2})^2 - \frac{1}{4}\right) = 16(x - \frac{1}{2})^2 - 4
   $$

#### For $$y$$:

1. Factor out the coefficient of $$y^2$$:
   $$
   24(y^2 + 2y)
   $$
2. Complete the square:
   $$
   y^2 + 2y = (y + 1)^2 - 1
   $$
   Thus,
   $$
   24\left((y + 1)^2 - 1\right) = 24(y + 1)^2 - 24
   $$

### Step 3: Substitute Back

Now substitute back into the equation:

$$
16(x - \frac{1}{2})^2 - 4 + 24(y + 1)^2 - 24 - 68 = 0
$$

Combine the constants:

$$
16(x - \frac{1}{2})^2 + 24(y + 1)^2 - 96 = 0
$$

Rearranging gives:

$$
16(x - \frac{1}{2})^2 + 24(y + 1)^2 = 96
$$

### Step 4: Divide by 96

To put it in standard form, divide everything by 96:

$$
\frac{16(x - \frac{1}{2})^2}{96} + \frac{24(y + 1)^2}{96} = 1
$$

This simplifies to:

$$
\frac{(x - \frac{1}{2})^2}{6} + \frac{(y + 1)^2}{4} = 1
$$

### Step 5: Identify Parameters

From the standard form $$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$, we identify:

- Center $$(h, k) = \left(\frac{1}{2}, -1\right)$$
- $$a^2 = 6$$ and $$b^2 = 4$$ (where $$a > b$$)

### Step 6: Calculate Vertices

The vertices are located at:

- $$ (h \pm a, k) $$
- $$ (h, k \pm b) $$

Calculating these:

- Vertices along the $$x$$-axis:
  - Smaller $$x$$-value: $$ \left(\frac{1}{2} - \sqrt{6}, -1\right) $$
  - Larger $$x$$-value: $$ \left(\frac{1}{2} + \sqrt{6}, -1\right) $$

- Vertices along the $$y$$-axis:
  - $$ \left(\frac{1}{2}, -1 - 2\right) = \left(\frac{1}{2}, -3\right) $$
  - $$ \left(\frac{1}{2}, -1 + 2\right) = \left(\frac{1}{2}, 1\right) $$

### Step 7: Calculate Foci

The foci are located at:

- $$ (h \pm c, k) $$ where $$ c = \sqrt{a^2 - b^2} = \sqrt{6 - 4} = \sqrt{2} $$

Calculating the foci:

- Smaller $$x$$-value: $$ \left(\frac{1}{2} - \sqrt{2}, -1\right) $$
- Larger $$x$$-value: $$ \left(\frac{1}{2} + \sqrt{2}, -1\right) $$

### Step 8: Calculate Eccentricity

The eccentricity $$e$$ is given by:

$$
e = \frac{c}{a} = \frac{\sqrt{2}}{\sqrt{6}} = \frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3}
$$

### Final Results

- Center: $$ (x, y) = \left( \frac{1}{2}, -1 \right) $$
- Vertices: 
  - $$ (x, y) = \left( \frac{1}{2} - \sqrt{6}, -1 \right) $$ (smaller $$x$$-value)
  - $$ (x, y) = \left( \frac{1}{2} + \sqrt{6}, -1 \right) $$ (larger $$x$$-value)
- Foci:
  - $$ (x, y) = \left( \frac{1}{2} - \sqrt{2}, -1 \right) $$ (smaller $$x$$-value)
  - $$ (x, y) = \left( \frac{1}{2} + \sqrt{2}, -1 \right) $$ (larger $$x$$-value)
- Eccentricity: $$ e = \frac{\sqrt{3}}{3} $$
Center: $$ \left( \frac{1}{2}, -1 \right) $$ Vertices: - Smaller $$ x $$-value: $$ \left( \frac{1}{2} - \sqrt{6}, -1 \right) $$ - Larger $$ x $$-value: $$ \left( \frac{1}{2} + \sqrt{6}, -1 \right) $$ Foci: - Smaller $$ x $$-value: $$ \left( \frac{1}{2} - \sqrt{2}, -1 \right) $$ - Larger $$ x $$-value: $$ \left( \frac{1}{2} + \sqrt{2}, -1 \right) $$ Eccentricity: $$ \frac{\sqrt{3}}{3} $$

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