Question

(x-4)^2/36 + (y+2)^2/16 =1
What is the sum of the focal radii?
What are the coordinates of the foci?

Answer 1

\[\frac{ (x-4)^2 }{ 36 } + \frac{ (y+2)^2 }{ 16 } =1\]

Answer 2

help please

Answer 3

\(\bf \cfrac{(x-4)^2}{36}+\cfrac{(y+2)^2}{16}=1\implies \cfrac{(x-4)^2}{6^2}+\cfrac{(y+2)^2}{4^2}=1\\ \quad \\
\implies \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\\ \quad \\ \quad \\
\textit{distance from the vertex to the foci is "c"}\quad c=\sqrt{a^2-b^2}\)

Answer 4

is that the answer to the first question

Answer 5

hmmm I assume by "focal radii" it's meant the sum of both distances from the vertex

Answer 6

I Have no clue honestly.. My teacher just gave me work and didn't even give me a lesson over this.

Answer 7

so all you really need to find is the "c" distance, to get the sum of them and their coordinates, notice that the "a" component is the BIGGER denominator and is under the "x" variable, thus that means the ellipse is of traverse horizontal axis, or a horizontal one

Answer 8

how do you find the coordinates? -_-

Answer 9

well, what did you get for "c"?

Answer 10

2?

Answer 11

2?

Answer 12

6-4 = 2..

Answer 13

\(\bf \cfrac{(x-4)^2}{\color{red}{36}}+\cfrac{(y+2)^2}{\color{red}{16}}=1\implies \cfrac{(x-4)^2}{6^2}+\cfrac{(y+2)^2}{4^2}=1\\ \quad \\
\implies \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\\ \quad \\ \quad \\
\textit{distance from the vertex to the foci is "c"}\quad c=\sqrt{\color{red}{a^2-b^2}}\)

Answer 14

notice, is the squared version of "a" and "b"

Answer 15

so 10? .. i dont know what im doing.. i thought 2 was right

Answer 16

well, a = 6, b = 4
\(\bf a^2= 36\quad b^2= 16\)

Answer 17

\(\bf \cfrac{(x-4)^2}{36}+\cfrac{(y+2)^2}{16}=1\\ \quad \\
c=\sqrt{a^2-b^2}\implies c=\sqrt{20}\implies c=2\sqrt{5}\)

Answer 18

wow.. im dumb.. lol thanks

Answer 19

notice the center of the parabola, at (h, k) \(\bf \cfrac{(x-4)^2}{36}+\cfrac{(y+2)^2}{16}=1\implies \cfrac{(x-(\color{red}{+4}))^2}{36}+\cfrac{(y-(\color{red}{-2}))^2}{16}=1\\ \quad \\
\implies \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\)

so your center is at ( 4, -2 )

is a horizontal ellipse, like
so the focii are at \(\bf (4 \pm 2\sqrt{5}\quad ,\quad -2)\)

Answer 20

and their sum would be \(\bf 2\sqrt{5}+2\sqrt{5}\) I'd think

Answer 21

thanks :) i also have to find the eccentricity of the ellipse.

Answer 22

\(\bf \textit{eccentricity of an ellipse}=\cfrac{c}{a}\)

Answer 23

oh okay gracias

Answer 24

yw