How to re-write this standard form equation in conic form.

Question

The question that is circled (2) is the question I need help with.

(https://i.stack.imgur.com/y3jMT.jpg)

Answer 1

Note that, expanding the squares and bringing everything to the left side, you get

$$ x^2 - 8x + 16 + y^2 + 64 - 16y - 10 = 0 $$ Compare the standard form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ with what you get: $$ 1 x^2 + 0 xy + 1y^2 - 8x - 16y + 70 = 0 $$

You can write this as a quadratic equation $\mathbf{x}^T Q \mathbf{x} = 0$, where $$ Q = \left( \begin{array}{c c c} 1 & 0 & -4 \\ 0 & 1 & -8 \\ -4 & -8 & 70 \end{array} \right) $$ and $\mathbf{x} = \left( \begin{array}{c} x \\ y \\ 1 \end{array} \right) $.

The quadratic equation $\mathbf{x}^T Q \mathbf{x}$ corresponding to the standard form generally has $$ Q = \left( \begin{array}{c c c} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{array} \right) $$ Make sure you take a moment to write it down analytically and understand why you can write it as such.