Question
What conic section is represented by the equation 3x^2+2y^2-24x+30=0? Explain how you know?
is the ellipse represented by the equation (x-5)^2/3 + (y+1)^2/5 =1 oriented horizontally or vertically? explain how you know
is the ellipse represented by the equation (x-5)^2/3 + (y+1)^2/5 =1 oriented horizontally or vertically? explain how you know
Answers
GPT-4o mini
To determine the conic section represented by the equation \( 3x^2 + 2y^2 - 24x + 30 = 0 \), we first rearrange it into a more standard form by completing the square.
1. **Rearranging the Equation**:
\[
3x^2 - 24x + 2y^2 + 30 = 0
\]
Move \( 30 \) to the right side:
\[
3x^2 - 24x + 2y^2 = -30
\]
2. **Completing the Square for \( x \)**:
Factor out \( 3 \) from the \( x \) terms:
\[
3(x^2 - 8x) + 2y^2 = -30
\]
Complete the square for \( x^2 - 8x \):
\[
x^2 - 8x = (x-4)^2 - 16
\]
Substitute back:
\[
3((x - 4)^2 - 16) + 2y^2 = -30
\]
This simplifies to:
\[
3(x - 4)^2 - 48 + 2y^2 = -30
\]
Combine and simplify:
\[
3(x - 4)^2 + 2y^2 = 18
\]
Divide the entire equation by \( 18 \):
\[
\frac{(x - 4)^2}{6} + \frac{y^2}{9} = 1
\]
This is in the standard form of an ellipse:
\[
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
\]
where \( (h, k) \) is the center and \( a^2 = 6, b^2 = 9 \). Since both \( a^2 \) and \( b^2 \) are positive and unequal, this confirms it is an ellipse.
Next, we identify the orientation based on the denominators:
- Since \( a^2 < b^2 \) (6 < 9), the major axis corresponds to the \( y \)-direction, meaning the ellipse is oriented vertically.
**Conclusion**: The conic section represented by the equation \( 3x^2 + 2y^2 - 24x + 30 = 0 \) is an ellipse, and it is oriented vertically in the equation \( \frac{(x-5)^2}{3} + \frac{(y+1)^2}{5} = 1 \).
1. **Rearranging the Equation**:
\[
3x^2 - 24x + 2y^2 + 30 = 0
\]
Move \( 30 \) to the right side:
\[
3x^2 - 24x + 2y^2 = -30
\]
2. **Completing the Square for \( x \)**:
Factor out \( 3 \) from the \( x \) terms:
\[
3(x^2 - 8x) + 2y^2 = -30
\]
Complete the square for \( x^2 - 8x \):
\[
x^2 - 8x = (x-4)^2 - 16
\]
Substitute back:
\[
3((x - 4)^2 - 16) + 2y^2 = -30
\]
This simplifies to:
\[
3(x - 4)^2 - 48 + 2y^2 = -30
\]
Combine and simplify:
\[
3(x - 4)^2 + 2y^2 = 18
\]
Divide the entire equation by \( 18 \):
\[
\frac{(x - 4)^2}{6} + \frac{y^2}{9} = 1
\]
This is in the standard form of an ellipse:
\[
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
\]
where \( (h, k) \) is the center and \( a^2 = 6, b^2 = 9 \). Since both \( a^2 \) and \( b^2 \) are positive and unequal, this confirms it is an ellipse.
Next, we identify the orientation based on the denominators:
- Since \( a^2 < b^2 \) (6 < 9), the major axis corresponds to the \( y \)-direction, meaning the ellipse is oriented vertically.
**Conclusion**: The conic section represented by the equation \( 3x^2 + 2y^2 - 24x + 30 = 0 \) is an ellipse, and it is oriented vertically in the equation \( \frac{(x-5)^2}{3} + \frac{(y+1)^2}{5} = 1 \).