MATH SOLVE

3 months ago

Q:
# PLEASE HELP8.03a1. Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1. A) Vertices: (± 5, 0); Foci: (± 3, 0) B) Vertices: (± 3, 0); Foci: (± 5, 0) C) Vertices: (0, ± 3); Foci: (0, ±5) D) Vertices: (0, ± 5); Foci: (0, ± 3)2. Find the vertices and foci of the hyperbola with equation quantity x minus three squared divided by sixteen minus the quantity of y plus four squared divided by nine = 1. A) Vertices: (7, -4), (-1, -4); Foci: (-2, -4), (8, -4) B) Vertices: (-4, 7), (-4, -1); Foci: (-4, -2), (-4, 8) C) Vertices: (6, -4), (0, -4); Foci: (0, -4), (6, -4) D) Vertices: (-4, 6), (-4, 0); Foci: (-4, 0), (-4, 6)3. Graph the hyperbola with equation A horizontal hyperbola is shown on the coordinate plane centered at, negative one, four, with vertices at, negative four, negative four and two, negative four. = 1. A) A vertical hyperbola is shown on the coordinate plane centered at the origin with vertices at, zero, two and zero, negative two. B) A horizontal hyperbola is shown on the coordinate plane centered at, negative three, three, with vertices at, negative five, three and negative 1, three. C) A vertical hyperbola is shown on the coordinate plane centered at, three, negative three, with vertices at, three, negative one and three, negative five. D) A horizontal hyperbola is shown on the coordinate plane centered at the origin with vertices at, negative two, zero and two, zero.4. Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7). A) y squared over forty nine minus x squared over four = 1 B) y squared over forty five minus x squared over four = 1 C) y squared over four minus x squared over forty nine = 1 D) y squared over four minus x squared over forty five = 15. Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ± three divided by four times x..

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