Question

find the vertices and foci of the ellipse given by the equation 4x2 + 9y2 = 144.
a.) vertices:(0,6),(0,-6);foci:(0,2\sqrt(5),(0,-2\sqrt(5)
b.)vertices:(4,0),(-4,0);foci:(2\sqrt(5},0),(-2\sqrt(5),0)
c.)vertices:(0,4),(0,-4);foci:(0,2\sqrt(5)),(0,-2\sqrt(5))
d,)vertices:(6,0),(-6,0);foci:(2\sqrt(5),0),(-2\sqrt(5),0)

Answer 1

turn to standard form
first divide by 144

Answer 2

\[\frac{x ^{2} }{36 }+\frac{y ^{2} }{16 }=1\]

Answer 3

Right, and that can be expressed as\[\frac{ x ^{2} }{ 6^{2} }+\frac{ y ^{2} }{ 4^{2} }=1\]

Answer 4

Look at the denominators. That 6 is greater than that 4, so we know the "major axis) of the ellips is horizonta (not verticaL), and has length 6. Yes, amoodarya, nice and helpful diagram.

Answer 5

\[a^{2}=36 ,b^{2}=16 ,c^{2}=a^{2}-b^{2}=36-16=20 \rightarrow c=\sqrt{20}\]

Answer 6

major axis=6
minor axis=4
How do you use this information to determine how far along the x axis the right-hand focus is from the center of the ellilpse?

Answer 7

amoodarya and I make a good team: he provides the equations, I interpret what we're doing.

Answer 8

yes we can do that

Answer 9

Amoodarya: How about drawing a new diagram showing how a, b and c are related, and what c has to do with the location of the focus? (I've forgotten)

Answer 10

Nice drawing. How are a, b and c related algebraically?