If the graph of `3x^2+4y^2-6x+8y-5=0` and `(x-2)^2=4(y+2)` are drawn on the same coordinate system, at how many points do they intersect?

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To determine the number of intersection points between the graphs of the equations \(3x^2 + 4y^2 - 6x + 8y - 5 = 0\) and \((x - 2)^2 = 4(y + 2)\), we will follow these steps:

### Step 1: Identify the conic sections
The first equation \(3x^2 + 4y^2 - 6x + 8y - 5 = 0\) represents an ellipse, while the second equation \((x - 2)^2 = 4(y + 2)\) represents a parabola.

### Step 2: Rewrite the ellipse equation
We will complete the square for the ellipse equation.

1. **Group the x and y terms:**
\[ 3(x^2 - 2x) + 4(y^2 + 2y) - 5 = 0 \]

2. **Complete the square for \(x\):**
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Thus,
\[ 3((x - 1)^2 - 1) = 3(x - 1)^2 - 3 \]

3. **Complete the square for \(y\):**
\[ y^2 + 2y = (y + 1)^2 - 1 \]
Thus,
\[ 4((y + 1)^2 - 1) = 4(y + 1)^2 - 4 \]

4. **Substituting back into the equation:**
\[ 3(x - 1)^2 - 3 + 4(y + 1)^2 - 4 - 5 = 0 \]
Simplifying gives:
\[ 3(x - 1)^2 + 4(y + 1)^2 - 12 = 0 \]
Rearranging results in:
\[ 3(x - 1)^2 + 4(y + 1)^2 = 12 \]
Dividing by 12:
\[ \frac{(x - 1)^2}{4} + \frac{(y + 1)^2}{3} = 1 \]
This is the standard form of an ellipse centered at \((1, -1)\).

### Step 3: Rewrite the parabola equation
The parabola equation \((x - 2)^2 = 4(y + 2)\) can be rewritten as:
\[ y + 2 = \frac{(x - 2)^2}{4} \]
Thus,
\[ y = \frac{(x - 2)^2}{4} - 2 \]

### Step 4: Find points of intersection
To find the points of intersection, we substitute the expression for \(y\) from the parabola into the ellipse equation.

1. Substitute \(y\):
\[ 3(x - 1)^2 + 4\left(\frac{(x - 2)^2}{4} - 2 + 1\right)^2 = 12 \]
Simplifying:
\[ 3(x - 1)^2 + 4\left(\frac{(x - 2)^2}{4} - 1\right)^2 = 12 \]
\[ 3(x - 1)^2 + (x - 2)^2 - 4 = 12 \]
\[ 3(x - 1)^2 + (x - 2)^2 = 16 \]

2. Expand and combine like terms:
\[ 3(x^2 - 2x + 1) + (x^2 - 4x + 4) = 16 \]
\[ 3x^2 - 6x + 3 + x^2 - 4x + 4 = 16 \]
\[ 4x^2 - 10x + 7 = 16 \]
\[ 4x^2 - 10x - 9 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ a = 4, \quad b = -10, \quad c = -9 \]
\[ D = b^2 - 4ac = (-10)^2 - 4 \cdot 4 \cdot (-9) = 100 + 144 = 244 \]
Since the discriminant \(D > 0\), there are 2 distinct real solutions for \(x\).

### Step 6: Conclusion
Since there are 2 distinct values for \(x\), we can substitute back to find the corresponding \(y\) values, resulting in 2 intersection points between the ellipse and the parabola.

### Final Answer
The graphs intersect at **2 points**.