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# 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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Conic sections
asked 2021-02-05
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 $$16x^2 + 64x - 9y^2 + 18y - 89 = 0$$

## Answers (1)

2021-02-06
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 $$\frac{(x+2)^2}{9} - \frac{(y - 1)^2}{16} = 1$$

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