# a) determine the type of conic b) find the standard form of the equation Parabolas: vertex, focus, directrix Circles: Center, radius Ellipses: center, vertices, co-vertices, foci Hyperbolas: center, vertices, co-vertices, foci, asymptotes 16x^2 + 64x - 9y^2 + 18y - 89 = 0

a) determine the type of conic b) find the standard form of the equation Parabolas: vertex, focus, directrix Circles: Center, radius Ellipses: center, vertices, co-vertices, foci Hyperbolas: center, vertices, co-vertices, foci, asymptotes $16{x}^{2}+64x-9{y}^{2}+18y-89=0$
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Alara Mccarthy

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 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 $16{x}^{2}+64x-9{y}^{2}+18y-89=0$
$16{x}^{2}+64x-9{y}^{2}+18y=89$ Factor out 16 from the x portion and -9 from the y portion $16\left({x}^{2}+4x\right)-9\left({y}^{2}-2y\right)=89$ Then complete square for each section and to balance right side add or subtract the same amount. $16\left({x}^{2}+4x+4\right)-9\left({y}^{2}-2y+1\right)=89+16\left(4\right)-9\left(1\right)$
$16\left(x+2{\right)}^{2}-9\left(y-1{\right)}^{2}=144$ Then divide both sides by 144 $\frac{16\left(x+2{\right)}^{2}}{144}-\frac{9\left(y-1{\right)}^{2}}{144}=\frac{144}{144}$
$\frac{\left(x+2{\right)}^{2}}{9}-\frac{\left(y-1{\right)}^{2}}{16}=1$ Answer $\frac{\left(x+2{\right)}^{2}}{9}-\frac{\left(y-1{\right)}^{2}}{16}=1$