Question

What are the foci of the ellipse given by the equation (x-5)^2/81 + (y+1)^2/9 =1

Answer 1

by convention; there are 3 variables used in an ellipse; a b and c

Answer 2

c is the measure of the foci from the center

a is the measure of the widest vertex from the center

and b is the measure of the shortest vertex from the center

Answer 3

\[\frac{(x-cx)^2}{a^2}+\frac{(y-cy)^2}{b^2}=1\]

Answer 4

err, the cs in that one are for centers ....

Answer 5

a b c form a rt triangle; with b and c being legs since the pythag thrm had already used a and b as legs they want to throw you for a loop here

Answer 6

therefore; b^2 + c^2 = a^2; or to rewrite it another way: c = sqrt(a^2-b^2)

Answer 7

since a^2 is the larger denom; we can form this up using the given equation as:

c = sqrt(81-9) = sqrt(72)

Answer 8

the bigger a^2 is under the x parts so we measure from the center parallel to (along) the x axis

Answer 9

given the equation parts: (x-5) and (y+1)
our center is the point ( 5 , -1 )

so lets add and subtract sqrt(72) from the x part to determine the focuses

(5-sqrt(72), -1)
and
(5+sqrt(72), -1)


simplify as wanted

Answer 10

thanks....can you answer another question please

Answer 11

What are the vertices of the ellipse given by the equation (x+6)^2/4 +(y+9)^2/1 =1