Question

21. What is the length of the major axis?




22. What is the length of the minor axis?



23. What is the sum of the focal radii?




24. What are the coordinates of the foci?

Answer 1

\[\frac{ (x+4)^{2} }{ 12 } \frac{ (y-1)^{2} }{ 8 } =1\]

Answer 2

help please

Answer 3

\(\bf \cfrac{(x+4)^2}{12}+\cfrac{(y-1)^2}{8}=1\implies \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\\ \quad \\
\cfrac{(x+4)^2}{(\sqrt{12})}+\cfrac{(y-1)^2}{(\sqrt{8})^2}=1\)

the major axis is "a" units from the center of the ellipse, thus a + a is how long it's

the minor is "b" units from the center, thus b + b is how long it's

the distance from the center to either focus is "c", \(\bf c=\sqrt{a^2-b^2}\) its sum would be c + c

notice the BIGGER denominator is under the "x" variable, thus that means is a horizontal ellipse like

the coordinates of the center are at (h, k)
the coordinates of either focus is at \(\bf ( h \pm c, k)\)

Answer 4

hmmm missed a squared... so \((\bf \cfrac{(x+4)^2}{12}+\cfrac{(y-1)^2}{8}=1\implies \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\\ \quad \\
\cfrac{(x+4)^2}{(\sqrt{12})^2}+\cfrac{(y-1)^2}{(\sqrt{8})^2}=1\)

Answer 5

oohh oops

Answer 6

keeping in mind that
\(\bf \cfrac{(x+4)^2}{12}+\cfrac{(y-1)^2}{8}=1\implies \cfrac{(x-(\color{red}{-4}))^2}{12}+\cfrac{(y-\color{red}{1})^2}{8}=1\)