Question

WILL GIVE BRAINLIEST, PLEASE HELP! Given the foci and vertices of an ellipse, complete the following steps.

Answer 1

Answer:

a) The location of the center of the ellipse is $(4,-6)$. It is determine by knowing the location of foci and Midpoint formula.

b) The standard equation of the ellipse is $\frac{(x-4)^{2}}{31}+\frac{(y+6)^{2}}{81} = 1$.

c) Major semiaxis ($a$): $a = 9$, Minor semiaxis ($b$): $b = \sqrt{31}$

Step-by-step explanation:

a) Foci are located in the major axis of the ellipse, given the locations of ends of the major axis and locations of the foci. We can determine the location of the center of the ellipse by Midpoint formula:

$(h,k) = \frac{1}{2}\cdot F_{1}(x,y) + \frac{1}{2}\cdot F_{2} (x,y)$ (1)

Where $F_{1} (x,y)$ and $F_{2} (x,y)$ are the locations of the foci.

If we know that $F_{1}(x,y) = (4, -6-5\sqrt{2})$ and $F_{2} (x,y) = (4,-6+5\sqrt{2})$, then the location of the center of the ellipse is:

$(h,k) = (4, -6)$

The location of the center of the ellipse is $(4,-6)$. It is determine by knowing the location of foci and Midpoint formula.

b) Now we need to determine the length of each semiaxis:

Major semiaxis ($a$)

$a = \sqrt{[V_{2}(x,y)-(h,k)]\,\bullet\,[V_{2}(x,y)-(h,k)]}$ (2)

$V_{2}(x,y)-(h,k) = (4,3)-(4,-6)$

$V_{2}(x,y) - (h,k) = (0, 9)$

$a = 9$

Please notice that $V_{2}(x,y)$ is one of the vertices of the ellipse in the major axis.

Distance between focus and center ($c$)

$c = \sqrt{[F_{2}(x,y)-(h,k)]\,\bullet\,[F_{2}(x,y)-(h,k)]}$

$F_{2}(x,y) -(h,k) = (4-4,-6+5\sqrt{2}+6)$

$F_{2}(x,y) - (h,k) = (0, 5\sqrt{2})$

$c = 5\sqrt{2}$

Minor semiaxis ($b$)

$c^{2} = a^{2}-b^{2}$ (3)

$b = \sqrt{a^{2}-c^{2}}$

If we know that $a = 9$ and $c = 5\sqrt{2}$, then the length of the minor semiaxis is:

$b = \sqrt{31}$

In this case, the standard equation of the ellipse is defined by this formula:

$\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}} = 1$

If we know that $a = 9$, $b = \sqrt{31}$ and $(h,k) = (4, -6)$, then the standard equation of the ellipse is:

$\frac{(x-4)^{2}}{31}+\frac{(y+6)^{2}}{81} = 1$

The standard equation of the ellipse is $\frac{(x-4)^{2}}{31}+\frac{(y+6)^{2}}{81} = 1$.

c) The lengths of the axes are, respectively:

Major semiaxis ($a$): $a = 9$

Minor semiaxis ($b$): $b = \sqrt{31}$

Answer 2

Answer:

a) The location of the center of the ellipse is . It is determine by knowing the location of foci and Midpoint formula.

b) The standard equation of the ellipse is .

c) Major semiaxis (): , Minor semiaxis ():

Step-by-step explanation: