Question

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Question
Identify the conic section represented by the equation 4 y squared minus 9 x squared minus 48 y minus 18 x plus 99 equals 0.
(1 point)
Responses

circle
circle

ellipse
ellipse

hyperbola
hyperbola

parabola

Answer 1

To identify the conic section represented by the equation \( 4y^2 - 9x^2 - 48y - 18x + 99 = 0 \), we can rearrange the equation and look for specific characteristics of the conic.

1. **Rearranging the equation:**
\[
4y^2 - 48y - 9x^2 - 18x + 99 = 0
\]

2. **Group the \(y\) terms and \(x\) terms:**
\[
4(y^2 - 12y) - 9(x^2 + 2x) + 99 = 0
\]

3. **Complete the square for the \(y\) terms:**
\[
y^2 - 12y = (y - 6)^2 - 36 \implies 4((y - 6)^2 - 36) = 4(y - 6)^2 - 144
\]

4. **Complete the square for the \(x\) terms:**
\[
x^2 + 2x = (x + 1)^2 - 1 \implies -9((x + 1)^2 - 1) = -9(x + 1)^2 + 9
\]

5. **Substituting back into the equation:**
\[
4(y - 6)^2 - 144 - 9(x + 1)^2 + 9 + 99 = 0
\]
Simplifying gives:
\[
4(y - 6)^2 - 9(x + 1)^2 - 36 = 0
\]
\[
4(y - 6)^2 - 9(x + 1)^2 = 36
\]

6. **Dividing by 36:**
\[
\frac{4(y - 6)^2}{36} - \frac{9(x + 1)^2}{36} = 1
\]
\[
\frac{(y - 6)^2}{9} - \frac{(x + 1)^2}{4} = 1
\]

This is the standard form of a hyperbola, which is given by the equation:
\[
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1
\]

where \(a^2 = 9\) and \(b^2 = 4\).

Thus, the correct response is:

**hyperbola**.