Questions: The intersection of the surface (9+x^2+fracy^24=z^2) and the plane (z=0) resembles which of the following?

$The intersection of the surface (9+x^2+fracy^24=z^2) and the plane (z=0) resembles which of the following?$

Transcript text: The intersection of the surface $9+x^{2}+\frac{y^{2}}{4}=z^{2}$. and the plane $z=0$ resembles which of the following? ? $\square$ - Part 4: Classify

Solution

Solution Steps

Step 1: Identify the given surface equation

The given surface equation is: \[ 9 + x^2 + \frac{y^2}{4} = z^2 \]

Step 2: Set the plane equation

The plane equation is: \[ z = 0 \]

Step 3: Substitute the plane equation into the surface equation

Substitute $ z = 0 $ into the surface equation: \[ 9 + x^2 + \frac{y^2}{4} = 0 \]

Step 4: Simplify the equation

Rearrange the equation: \[ x^2 + \frac{y^2}{4} = -9 \]

Step 5: Analyze the simplified equation

The equation $ x^2 + \frac{y^2}{4} = -9 $ represents an ellipse, but since the right-hand side is negative, there are no real solutions. This means the intersection does not exist in the real plane.

Final Answer

The intersection of the surface $ 9 + x^2 + \frac{y^2}{4} = z^2 $ and the plane $ z = 0 $ does not exist in the real plane. Therefore, the correct answer is the graph that shows no intersection, which is option F.

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