11. [-/6 Points] DETAILS MY NOTES LARPCALC10 10.3.039. Find the center, vertices, foci, and eccentricity of the ellipse. $$ 16 x^{2}+4 y^{2}+64 x-32 y+64=0 $$ center $$ (x, y)=(\square) $$ vertices $$ \quad(x, y)=(\square) $$ (smaller $$ y $$-value) $$ (x, y)=(\square)(\text { larger } y \text {-value) } $$ foci $$ \quad(x, y)=( $$ $$ \square $$ ) (smaller $$ y $$-value) $$ (x, y)=(\square) \text { (larger } y \text {-value) } $$ eccentricity $$ \square $$ Sketch the ellipse.

Question

UpStudy AI · Accepted Answer

We start with the given ellipse:
$$
16x^2+4y^2+64x-32y+64=0.
$$

**Step 1. Group the terms and complete the square.**

Group the $$x$$ and $$y$$ terms:
$$
16(x^2+4x)+4(y^2-8y)+64=0.
$$

Complete the square for the $$x$$–terms. For $$x^2+4x$$, take half of $$4$$ (which is $$2$$) and square it to get $$4$$:
$$
x^2+4x = (x+2)^2-4.
$$
Thus,
$$
16\bigl[(x+2)^2-4\bigr]=16(x+2)^2-64.
$$

Now complete the square for the $$y$$–terms. For $$y^2-8y$$, take half of $$-8$$ (which is $$-4$$) and square it to get $$16$$:
$$
y^2-8y = (y-4)^2-16.
$$
Thus,
$$
4\bigl[(y-4)^2-16\bigr]=4(y-4)^2-64.
$$

Substitute these into the original equation:
$$
16(x+2)^2-64+4(y-4)^2-64+64=0.
$$

Combine the constants:
$$
16(x+2)^2+4(y-4)^2-64=0.
$$

Add $$64$$ to both sides:
$$
16(x+2)^2+4(y-4)^2=64.
$$

**Step 2. Write the equation in standard form.**

Divide the entire equation by $$64$$:
$$
\frac{16(x+2)^2}{64}+\frac{4(y-4)^2}{64}=1.
$$
Simplify:
$$
\frac{(x+2)^2}{4}+\frac{(y-4)^2}{16}=1.
$$

This is the standard form of an ellipse where the center is at $$(-2,4)$$.

**Step 3. Identify the center, vertices, foci, and eccentricity.**

- **Center:**  
  The center is given by the translation:
  $$
  (x,y)=(-2,4).
  $$

- **Vertices:**  
  Since the denominator under $$(y-4)^2$$ is larger ($$16>4$$), the major axis is vertical. The standard form tells us:
  $$
  a^2=16 \quad\Rightarrow\quad a=4,\qquad b^2=4 \quad\Rightarrow\quad b=2.
  $$
  The vertices lie $$a$$ units above and below the center along the $$y$$–axis:
  $$
  (-2,4-4)=(-2,0) \quad\text{and}\quad (-2,4+4)=(-2,8).
  $$
  The vertex with the smaller $$y$$–value is $$(-2,0)$$ and with the larger $$y$$–value is $$(-2,8)$$.

- **Foci:**  
  For an ellipse with a vertical major axis, the foci are located along the vertical line through the center. The distance $$c$$ is found from:
  $$
  c^2=a^2-b^2=16-4=12\quad\Rightarrow\quad c=2\sqrt{3}.
  $$
  Thus, the foci are:
  $$
  (-2,4-2\sqrt{3}) \quad\text{and}\quad (-2,4+2\sqrt{3}).
  $$
  The focus with the smaller $$y$$–value is $$(-2,4-2\sqrt{3})$$ and with the larger $$y$$–value is $$(-2,4+2\sqrt{3})$$.

- **Eccentricity:**  
  The eccentricity is given by:
  $$
  e=\frac{c}{a}=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}.
  $$

**Summary of Answers:**

- Center: $$(x,y)=(-2,\,4)$$
- Vertices:  
  $$\quad (x,y)=(-2,\,0)$$ (smaller $$y$$–value)  
  $$\quad (x,y)=(-2,\,8)$$ (larger $$y$$–value)
- Foci:  
  $$\quad (x,y)=(-2,\,4-2\sqrt{3})$$ (smaller $$y$$–value)  
  $$\quad (x,y)=(-2,\,4+2\sqrt{3})$$ (larger $$y$$–value)
- Eccentricity: $$\frac{\sqrt{3}}{2}$$
Center: $$(-2, 4)$$ Vertices: $$(-2, 0)$$ (smaller $$y$$-value) $$(-2, 8)$$ (larger $$y$$-value) Foci: $$(-2, 4 - 2\sqrt{3})$$ (smaller $$y$$-value) $$(-2, 4 + 2\sqrt{3})$$ (larger $$y$$-value) Eccentricity: $$\frac{\sqrt{3}}{2}$$

Solución de inteligencia artificial de Upstudy

Responder

Solución

preguntas relacionadas

Latest Pre Calculus Questions