Question

write an equation of the ellipse with foci (0,+-2) and co-vertices at (+-1,0)

Answer 1

so the co-vertices, that is the "minor axis" has vertices at (1, 0) and (-1, 0)
notice the "y" coordinate didn't change, thus the minor is moving about the x-axis
that means the major is moving about the y-axis

the foci are at (0, 2) and (0, -2)
up 2 units, down 2 units

so obviously the center is the origin here

so what's the "a" component for the major axis moving over the y-axis?
well we dunno
we but we know that the distance "c" is the distance from either focus to the center
if the focus are at (0,+-2) , the obviously our "c = 2"

and we also know that in an ellipse c^2 = a^2 - b^2 <--- notice is minus

the minor axis vertices are at (1, 0) and (-1, 0), half that distance is our "b"
thus b = 1
c = 2
\(\bf c^2 = a^2 - b^2 \implies \sqrt{c^2+b^2} = a\\
a = \sqrt{4+1} \implies \sqrt{5}\\
a^2 = 5\)

Answer 2

so the answer would be x^2/5+y^2=1

Answer 3

?

Answer 4

recall, the major traverse axis is "y", not "x"
and the bigger denominator goes where the major axis is, in this case, the "y" fraction
\(\bf \cfrac{x^2}{1}+\cfrac{y^2}{5}=1\)

Answer 5

that is for an ellipse

Answer 6

what is the equation of a hyperbola with a=3 and c=7? assume that the transverse axis is horizontal

Answer 7

for a hyperbola, c^2 = a^2 + b^2 <--- notice is plus

so to get the "b" component, just \(\bf c^2 = a^2 + b^2 \implies b^2 = c^2 -a^2\\
b^2 = 7^2 -3^2 \implies b^2 = 40\)

Answer 8

x^2/40-y^2/p=1

Answer 9

?

Answer 10

well, technically yes and no, you see, in an ellipse "a" is always the bigger denominator
in a hyperbola, a or b can be smaller or bigger than the other

Answer 11

or is it x^2/9-y^2/40=1

Answer 12

so we know a, and we know b here, but who goes where.... I'm thinking, without extra checking though that the "x" will have the "a" component

Answer 13

or is it x^2/9-y^2/40=1

Answer 14

try instead -> x^2/40-y^/9=1

Answer 15

write the equation of the ellipse with its center at the origin. the height of the ellipse is 16 and the width is 8 units

Answer 16

hmm, extra checking seems that the "a" component in a hyperbola is ALWAYS on the POSITIVE fraction, so, "x" in this case, since the positive fraction is the one with the traverse axis

Answer 17

height is 16, a = 16/2 = 8
width is 8, b = 8/2 = 4

Answer 18

is taller than it's wide
meaning is vertical, thus the "a" or bigger denominator goes to the "y" fraction

Answer 19

x^2/8+y^2/4=1

Answer 20

well, a = 8, a^2 = 64
b^2 = 16

Answer 21

x^2/64+y^2/16=1

Answer 22

x^2/16+y^2/64=1 rather

it has a major vertical axis, thus the "y" keeps the bigger denominator