Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.
a) Since the coefficients of \(x^2 and y^2\) are of opposite signs. 16 and -9.
So it is an equation of hyperbola.
Answer(a): Hyperbola
b) To find the standard form of the equation we have to complete squares for x and y portions separately.
Add 89 to both sides
\(16x^2 + 64x - 9y^2 + 18y - 89 = 0\)

\(16x^2 + 64x - 9y^2 + 18y = 89\) Factor out 16 from the x portion and -9 from the y portion \(16(x^2 + 4x) - 9(y^2 - 2y) = 89\) Then complete square for each section and to balance right side add or subtract the same amount. \(16(x^2 + 4x + 4) - 9(y^2 - 2y + 1) = 89 + 16 (4) - 9(1)\)

\(16(x + 2)^2 - 9(y - 1)^2 = 144\) Then divide both sides by 144 \(\frac{16(x + 2)^2}{144} - \frac{9(y - 1)^2}{144} = \frac{144}{144}\)

\(\frac{(x+2)^2}{9} - \frac{(y - 1)^2}{16} = 1\) Answer \(\frac{(x+2)^2}{9} - \frac{(y - 1)^2}{16} = 1\)

\(16x^2 + 64x - 9y^2 + 18y = 89\) Factor out 16 from the x portion and -9 from the y portion \(16(x^2 + 4x) - 9(y^2 - 2y) = 89\) Then complete square for each section and to balance right side add or subtract the same amount. \(16(x^2 + 4x + 4) - 9(y^2 - 2y + 1) = 89 + 16 (4) - 9(1)\)

\(16(x + 2)^2 - 9(y - 1)^2 = 144\) Then divide both sides by 144 \(\frac{16(x + 2)^2}{144} - \frac{9(y - 1)^2}{144} = \frac{144}{144}\)

\(\frac{(x+2)^2}{9} - \frac{(y - 1)^2}{16} = 1\) Answer \(\frac{(x+2)^2}{9} - \frac{(y - 1)^2}{16} = 1\)