12. [-/6 Points] DETAILS MY NOTES LARPCALC10 10.3.040. Find the center, vertices, foci, and eccentricity of the ellipse. $$ \begin{array}{l} 9 x^{2}+4 y^{2}-90 x+16 y+97=0 \ \text { center } \quad(x, y)=(\square) \ \text { vertices } \quad(x, y)=(\square) \text { (smaller } y \text {-value) } \ (x, y)=(\square) \text { (larger } y \text {-value) } \ \text { foci } \quad(x, y)=(\square) \text { (smaller } y \text {-value) } \ (x, y)=(\square) \text { (larger } y \text {-value) } \ \text { eccentricity } \end{array} $$

Question

UpStudy AI · Accepted Answer

The given ellipse is
$$
9x^2 + 4y^2 - 90x + 16y + 97 = 0.
$$
We'll complete the square in $$x$$ and $$y$$.

1. Group the $$x$$ and $$y$$ terms:
$$
9x^2 - 90x + 4y^2 + 16y = -97.
$$

2. Factor out the coefficients:
$$
9(x^2 - 10x) + 4(y^2 + 4y) = -97.
$$

3. Complete the square for each group.

For $$x^2 - 10x$$:
   $$
   x^2 - 10x = (x-5)^2 - 25.
   $$
   
   For $$y^2+4y$$:
   $$
   y^2+4y = (y+2)^2 - 4.
   $$

4. Substitute back into the equation:
$$
9[(x-5)^2 - 25] + 4[(y+2)^2 - 4] = -97.
$$
Expand and simplify:
$$
9(x-5)^2 - 225 + 4(y+2)^2 - 16 = -97.
$$
$$
9(x-5)^2 + 4(y+2)^2 = -97 + 225 + 16 = 144.
$$

5. Divide the entire equation by $$144$$ to write it in standard form:
$$
\frac{9(x-5)^2}{144} + \frac{4(y+2)^2}{144} = 1.
$$
Simplify:
$$
\frac{(x-5)^2}{16} + \frac{(y+2)^2}{36} = 1.
$$

6. Identify the parameters:
   - Center: $$(5,\,-2)$$.
   - Denominators show $$a^2 = 36$$ and $$b^2 = 16$$. Since $$a^2 > b^2$$, the major axis is vertical.
   - Semi-major axis: $$a = 6$$.
   - Semi-minor axis: $$b = 4$$.

7. Find the vertices (along the major axis, vertical direction):
   - Vertices: $$(5,\, -2 \pm 6)$$.
   - Smaller $$y$$-value: $$(5,\, -8)$$.
   - Larger $$y$$-value: $$(5,\, 4)$$.

8. Find the foci (located along the major axis). First, compute:
$$
c^2 = a^2 - b^2 = 36 - 16 = 20,\quad c = \sqrt{20} = 2\sqrt{5}.
$$
Foci: $$(5,\, -2 \pm 2\sqrt{5})$$.
   - Smaller $$y$$-value: $$(5,\, -2 - 2\sqrt{5})$$.
   - Larger $$y$$-value: $$(5,\, -2 + 2\sqrt{5})$$.

9. Compute the eccentricity:
$$
e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}.
$$

Thus, the answers are:

- Center: $$(5,\,-2)$$
- Vertices: $$(5,\, -8)$$ (smaller $$y$$-value) and $$(5,\, 4)$$ (larger $$y$$-value)
- Foci: $$(5,\,-2-2\sqrt{5})$$ (smaller $$y$$-value) and $$(5,\,-2+2\sqrt{5})$$ (larger $$y$$-value)
- Eccentricity: $$\frac{\sqrt{5}}{3}$$
Center: (5, -2) Vertices: (5, -8) and (5, 4) Foci: (5, -2 - 2√5) and (5, -2 + 2√5) Eccentricity: √5/3

[-/6 Points]
DETAILS
MY NOTES
LARPCALC10 10.3.040.

Find the center, vertices, foci, and eccentricity of the ellipse.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Geometry Questions