Question

What are the vertices of the ellipse given by the equation the quantity x plus 3 end quantity squared over 16 plus the quantity y minus 4 end quantity squared over 121= 1?

Answer 1

(–3, 8) and (–3, 0)

(–3, –7) and (–3, 15)

(–14, 4) and (8, 4)

(1, 4) and (–7, 4)

Answer 2

http://www.clausentech.com/lchs/dclausen/algebra2/ellipses.htm

Answer 3

so \(\bf \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\\
\cfrac{(x+3)^2}{16}+\cfrac{(y-4)^2}{121} = 1 \implies \cfrac{(x+3)^2}{4^2}+\cfrac{(y-4)^2}{11^2} = 1\)

what do you think is the (h, k) point for the center of the ellipse?

Answer 4

h=3 and k=4 right?

Answer 5

well, keep in mind that (x+3) => ( x - ( - 3) )

Answer 6

and (y - 4) => ( y - ( +4 ) )

Answer 7

so h= 3 and k= -4?

Answer 8

the standard equation for an ellipse is \(\bf \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\)

Answer 9

right so where do i go from there

Answer 10

well, what's your center?

Answer 11

\(\bf \cfrac{(x+3)^2}{4^2}+\cfrac{(y-4)^2}{11^2} = 1
\implies \cfrac{(x-(-3))^2}{4^2}+\cfrac{(y-(+4))^2}{11^2} = 1\\
\cfrac{(x-\color{red}{h})^2}{a^2}+\cfrac{(y-\color{red}{k})^2}{b^2}=1\)

Answer 12

3,-4?

Answer 13

notice the 2nd version

Answer 14

-3, 4? sorry i dont follow

Answer 15

the standard equation is (x-h)
if you have (x + 3) and the "x" component would be 3, only if the standard equation would use (x +h), but it doesn't, it uses (x-h)
that means that for (x+3) to turn into a (x - h) form, it has to be ( x - ( -h) )
minus times minus = plus

Answer 16

so the center is at (-3, 4) = ( h, k)

the "a" component of an ellipse is the bigger SQUARED denominator
and it's also the major axis, and the vertices will be THAT MUCH FAR FROM THE center of the ellipse

whichever variable that has that bigger denominator, is where the ellipse is LONGER at
in your case, "y"