Find the standard form of the equation of the ellipse satisf

Find the standard form of the equation of the ellipse satisfying the given conditions. foci: (0,-4), (0, 4); vertices: (0,-5), (0, 5) A) x^2/16 + y^2/9 = 1 B) x^2/25 + y^2/9 = 1 C) x^2/16 + y^2/25 = 1 D) x^2/9 + y^2/25 = 1 Endpoints of major axis: (-2, -1) and (-2,7): endopoints of minor axis: (-4,3) and (0,3): A) (x - 3)^2/4 + (y+ 2)^2/16 = 1 B) (x - 2)^2/4 + (y - 3)^2/16 = 1 Convert the equation to the standard form for an ellipse by completing the square on x and y. 4x^2 + 16y^2 - 16x - 96y + 96 = 0 A) (x - 2)^2/4 + (y - 3)^2/16 = 1 B) (x - 2)^2/16 + (y - 3)^2/4 = 1 C) (x + 2)^2/16 + (y + 3)^2/4 = 1 D) (x - 3)62/16 + (y - 2)^2/4 = 1 Find the vertices and locate the foci for the hyperbola whose equation is given. y^2/25 - x^2/49 = 1 A) vertices: (0,-5), (0,5) foci: (- Squareroot74, 0), (Squareroot74, 0) B) vertices: (-5, 0), (5, 0) foci: (-7, 0), (7, 0) C) vertices: (0, - 5), (0, 5) foci: (0, - Squareroot74), (0, Squareroot74) D) vertices: (-7, 0), (7, 0) foci: (-7 Squareroot74, 0), (Squareroot74, 0)

Solution

1) foci ( 0,-4) and (0,4)

vertices ( 0,-5) and (0,5)

standard equation of ellipse is

y^2 / a^2 + x^2 / b^2 = 1

major axis 2a = 5+5 = 10

a = 5

a^2 = 25

c = 3

c^2 = 9

b^2 = a^2 - c^2

b^2 = 25 - 9 = 16

hence equation of ellipse is

(y^2/25) + (x^2/16) = 1

option c is correct

2) endpoints of major axis (-2,-1) and (-2,7)

endpoints of minor axis (-4,3) and (0,3)

calculating distance of major axis by applying distance formula

major axis length 2a = 8

a = 4

distance of minor axis 2b = 4

b = 2

centre is given by midpoint of major or minor axis which is (-2,3)

equation of ellipse is

(y-3)^2 / 16 + (x+2)^2 / 4 = 1 (option d )