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 16x2+64x−9y2+18y−89=0

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Alara Mccarthy · Accepted Answer

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 x2 and y2 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 16x2+64x−9y2+18y−89=0 16x2+64x−9y2+18y=89 Factor out 16 from the x portion and -9 from the y portion 16(x2+4x)−9(y2−2y)=89 Then complete square for each section and to balance right side add or subtract the same amount. 16(x2+4x+4)−9(y2−2y+1)=89+16(4)−9(1) 16(x+2)2−9(y−1)2=144 Then divide both sides by 144 16(x+2)2144−9(y−1)2144=144144 (x+2)29−(y−1)216=1 Answer (x+2)29−(y−1)216=1

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

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