Solved on Feb 06, 2024

Find the focus of the ellipse with equation $\frac{(x+4)^{2}}{16}+\frac{(y-3)^{2}}{20}=1$ .

STEP 1

Assumptions
1. The given equation is of an ellipse in standard form.
2. The standard form of an ellipse with horizontal major axis is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ , where $(h,k)$ is the center of the ellipse, $a$ is the semi-major axis, and $b$ is the semi-minor axis.
3. The standard form of an ellipse with vertical major axis is $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$ , where $(h,k)$ is the center of the ellipse, $a$ is the semi-major axis, and $b$ is the semi-minor axis.
4. The foci of an ellipse are located along the major axis, at a distance of $c$ from the center, where $c^2 = a^2 - b^2$ .
5. We are to calculate one of the foci of the given ellipse.

STEP 2

Identify the center $(h,k)$ , the values of $a^2$ and $b^2$ , and determine which axis is the major axis by comparing $a^2$ and $b^2$ .
$\text{Given equation: } \frac{(x+4)^{2}}{16}+\frac{(y-3)^{2}}{20}=1$
$\text{Comparing with standard form: } \begin{cases} h = -4 \\ k = 3 \\ a^2 = 20 \\ b^2 = 16 \end{cases}$

STEP 3

Determine which axis is the major axis by comparing $a^2$ and $b^2$ .
Since $a^2 = 20$ and $b^2 = 16$ , and $a^2 > b^2$ , the major axis is vertical.

STEP 4

Calculate the distance $c$ from the center to the foci using the relationship $c^2 = a^2 - b^2$ .
$c^2 = a^2 - b^2$
$c^2 = 20 - 16$

STEP 5

Calculate the value of $c$ .
$c^2 = 4$
$c = \sqrt{4}$
$c = 2$

STEP 6

Since the major axis is vertical, the foci are located at $(h, k \pm c)$ .
$\text{Foci: } (h, k \pm c)$
$\text{Foci: } (-4, 3 \pm 2)$

STEP 7

Write the coordinates of the two foci.
$\text{Focus 1: } (-4, 3 + 2) = (-4, 5)$
$\text{Focus 2: } (-4, 3 - 2) = (-4, 1)$
You can choose either focus as the solution. For instance, one of the foci of the ellipse is $(-4, 5)$ .

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Find the focus of the ellipse with equation $\frac{(x+4)^{2}}{16}+\frac{(y-3)^{2}}{20}=1$ .

STEP 1

STEP 2

STEP 3

Determine which axis is the major axis by comparing $a^2$ and $b^2$ .
Since $a^2 = 20$ and $b^2 = 16$ , and $a^2 > b^2$ , the major axis is vertical.

STEP 4

Calculate the distance $c$ from the center to the foci using the relationship $c^2 = a^2 - b^2$ .
$c^2 = a^2 - b^2$
$c^2 = 20 - 16$

STEP 5

Calculate the value of $c$ .
$c^2 = 4$
$c = \sqrt{4}$
$c = 2$

STEP 6

Since the major axis is vertical, the foci are located at $(h, k \pm c)$ .
$\text{Foci: } (h, k \pm c)$
$\text{Foci: } (-4, 3 \pm 2)$

STEP 7

Write the coordinates of the two foci.
$\text{Focus 1: } (-4, 3 + 2) = (-4, 5)$
$\text{Focus 2: } (-4, 3 - 2) = (-4, 1)$
You can choose either focus as the solution. For instance, one of the foci of the ellipse is $(-4, 5)$ .

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Find the focus of the ellipse with equation (x+4)216+(y−3)220=1\frac{(x+4)^{2}}{16}+\frac{(y-3)^{2}}{20}=116(x+4)2​+20(y−3)2​=1.

STEP 1

STEP 2

STEP 3

Determine which axis is the major axis by comparing a2a^2a2 and b2b^2b2.Since a2=20a^2 = 20a2=20 and b2=16b^2 = 16b2=16, and a2>b2a^2 > b^2a2>b2, the major axis is vertical.

STEP 4

Calculate the distance ccc from the center to the foci using the relationship c2=a2−b2c^2 = a^2 - b^2c2=a2−b2.c2=a2−b2 c^2 = a^2 - b^2 c2=a2−b2c2=20−16 c^2 = 20 - 16 c2=20−16

STEP 5

Calculate the value of ccc.c2=4 c^2 = 4 c2=4c=4 c = \sqrt{4} c=4​c=2 c = 2 c=2

STEP 6

Since the major axis is vertical, the foci are located at (h,k±c)(h, k \pm c)(h,k±c).Foci: (h,k±c) \text{Foci: } (h, k \pm c) Foci: (h,k±c)Foci: (−4,3±2) \text{Foci: } (-4, 3 \pm 2) Foci: (−4,3±2)

STEP 7

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Find the focus of the ellipse with equation $\frac{(x+4)^{2}}{16}+\frac{(y-3)^{2}}{20}=1$ .

Determine which axis is the major axis by comparing $a^2$ and $b^2$ .
Since $a^2 = 20$ and $b^2 = 16$ , and $a^2 > b^2$ , the major axis is vertical.

Calculate the distance $c$ from the center to the foci using the relationship $c^2 = a^2 - b^2$ .
$c^2 = a^2 - b^2$
$c^2 = 20 - 16$

Calculate the value of $c$ .
$c^2 = 4$
$c = \sqrt{4}$
$c = 2$

Since the major axis is vertical, the foci are located at $(h, k \pm c)$ .
$\text{Foci: } (h, k \pm c)$
$\text{Foci: } (-4, 3 \pm 2)$