Let's solve the problem step by step: ### Step 1: Write the given equation The given equation is: \[ 16x^2 + 9y^2 + 16z^2 - 32x - 36y + 36 = 0 \] ### Step 2: Group the terms and complete the square for \(x\) and \(y\) #### For \(x\): \[ 16x^2 - 32x = 16(x^2 - 2x) \] Complete the square: \[ x^2 - 2x = (x-1)^2 - 1 \] So, \[ 16(x^2 - 2x) = 16[(x-1)^2 - 1] = 16(x-1)^2 - 16 \] #### For \(y\): \[ 9y^2 - 36y = 9(y^2 - 4y) \] Complete the square: \[ y^2 - 4y = (y-2)^2 - 4 \] So, \[ 9(y^2 - 4y) = 9[(y-2)^2 - 4] = 9(y-2)^2 - 36 \] #### For \(z\): \[ 16z^2\] is already in the required form. ### Step 3: Substitute back and simplify Substitute the completed squares back into the equation: \[ 16(x-1)^2 - 16 + 9(y-2)^2 - 36 + 16z^2 + 36 = 0 \] Combine constants: \[ 16(x-1)^2 + 9(y-2)^2 + 16z^2 - 16 - 36 + 36 = 0 \] \[ 16(x-1)^2 + 9(y-2)^2 + 16z^2 - 16 = 0 \] \[ 16(x-1)^2 + 9(y-2)^2 + 16z^2 = 16 \] Divide both sides by 16: \[ \frac{(x-1)^2}{1} + \frac{(y-2)^2}{16/9} + \frac{z^2}{1} = 1 \] Or, \[ \frac{(x-1)^2}{1} + \frac{(y-2)^2}{(4/3)^2} + \frac{z^2}{1} = 1 \] ### Step 4: Identify the surface This is the standard form of an **ellipsoid**: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} + \frac{(z-l)^2}{c^2} = 1 \] where \((h,k,l)\) is the center, and \(a,b,c\) are the axes. So, the answer is: **ellipsoid** ### Step 5: Match the surface to the correct image Between the two images: - The left image shows a **double cone** (elliptic cone). - The right image shows an **ellipsoid**. So, the correct answer is the **right image**. --- ### **Final Answers:** 1. **ellipsoid** (third option) 2. **Right image** (second image) --- **Explanation:** - Completing the square brought the given equation to the standard form of an ellipsoid. - The right image matches the shape of an ellipsoid.