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Home » Thus, the radius of the base of the cone is \(\boxed4\). - Dyverse

Thus, the radius of the base of the cone is \(\boxed4\). - Dyverse

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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Complete the square for the \( x \)-terms:
9(x^2 - 4x) = 9((x-2)^2 - 4) = 9(x-2)^2 - 36
Complete the square for the \( y \)-terms:
4(y^2 + 4y) = 4((y+2)^2 - 4) = 4(y+2)^2 - 16
Substitute these back into the equation:
9(x-2)^2 - 36 + 4(y+2)^2 - 16 = -36
9(x-2)^2 + 4(y+2)^2 - 52 = -36
Add 52 to both sides:
9(x-2)^2 + 4(y+2)^2 = 16
Divide the entire equation by 16 to get the standard form:

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We use the vector triple product identity. From $ \mathbf{v} \times \m...
\mathbf{a} \times (\mathbf{v} \times \mathbf{a}) = \mathbf{a} \times \...
Using the identity $ \mathbf{a} \times (\mathbf{v} \times \mathbf{a})...
\|\mathbf{a}\|^2 \mathbf{v} = \mathbf{a} \times \mathbf{b}
Compute $ \|\mathbf{a}\|^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 $
Now compute $ \mathbf{a} \times \mathbf{b} $:
\mathbf{a} \times \mathbf{b} = \begin{vmatrix}
& -2 & 1 \\
\end{vmatrix} = \mathbf{i}(2 \cdot 1 - 3 \cdot (-2)) - \mathbf{j}(1 \c...
= \mathbf{i}(2 + 6) - \mathbf{j}(1 - 3) + \mathbf{k}(-2 - 2) = \langle...

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