SOLUTION: (x+4)^2/9+(y+3)^2 /4 =1

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Question 1209794: (x+4)^2/9+(y+3)^2 /4 =1


Found 3 solutions by CPhill, greenestamps, ikleyn:
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
First, I'll generate some data for plotting the graph.
```python?code_reference&code_event_index=3
import pandas as pd
import numpy as np
# Generate x values from -7 to -1 with a step of 0.1
x = np.arange(-7, -1, 0.1)
# Calculate y values for the positive and negative square roots
y_pos = -3 + 2 * np.sqrt(1 - (x + 4)**2 / 9)
y_neg = -3 - 2 * np.sqrt(1 - (x + 4)**2 / 9)
# Combine x and y values into a DataFrame
df = pd.DataFrame({'x': np.concatenate((x, x)), 'y': np.concatenate((y_pos, y_neg))})
# Display the first 5 rows
print(df.head().to_markdown(index=False, numalign="left", stralign="left"))
# Print the column names and their data types
print(df.info())
```
```text?code_stdout&code_event_index=3
| x | y |
|:-----|:---------|
| -7 | -3 |
| -6.9 | -2.48792 |
| -6.8 | -2.28198 |
| -6.7 | -2.12822 |
| -6.6 | -2.00222 |

RangeIndex: 120 entries, 0 to 119
Data columns (total 2 columns):
# Column Non-Null Count Dtype
--- ------ -------------- -----
0 x 120 non-null float64
1 y 120 non-null float64
dtypes: float64(2)
memory usage: 2.0 KB
None
```
Now, I'll use the generated data to plot the graph.
```python?code_reference&code_event_index=7
import altair as alt
# Create the Altair plot
chart = alt.Chart(df).mark_circle().encode(
x=alt.X('x', title='x'), # Set x-axis title
y=alt.Y('y', title='y'), # Set y-axis title
tooltip=['x', 'y'] # Add tooltips for interactivity
).properties(
title='Graph of (x+4)^2/9 + (y+3)^2/4 = 1' # Set plot title
).interactive() # Enable zooming and panning
# Save the plot
chart.save('graph_plot.json')
```
[json-tag: code-generated-json-0-1741108966998394562]
Here is the graph of the given equation $(x+4)^2/9+(y+3)^2 /4 =1$.
Let me know if you'd like to explore more about this equation or need further assistance with plotting other graphs.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


%28x%2B4%29%5E2%2F9%2B%28y%2B3%29%5E2+%2F4+=1

The equation has x^2 and y^2 terms and no xy term, so the equation is of an ellipse whose axes are horizontal and vertical.

The general equations are...

(1) major axis horizontal: %28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1

(2) major axis vertictal: %28x-h%29%5E2%2Fb%5E2%2B%28y-k%29%5E2%2Fa%5E2=1

In both forms, the center of the ellipse is (h,k), a is the semi-major axis, and b is the semi-minor axis.

Compare the given equation to those two general forms to find that the ellipse has a horizontal major axis and has...
center (-4,-3)
semi-major axis 3
semi-minor axis 2

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

It looks like the Artificial Intelligence (represented by @CPhill) reached a dead end and cannot answer correctly.

Meanwhile, the answer is simple:

This equation represents an ellipse with the center at the point (-4,-3)
with a major semi-axis of sqrt%289%29 = 3 units (horizontal) and a minor semi-axis of sqrt%284%29 = 2 units (vertical).