QuestionDetermine the properties of the ellipse defined by $\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$ .

Studdy Solution

STEP 1

Assumptions1. The given equation is of an ellipse in standard form, $\frac{x^{}}{a^{}}+\frac{y^{}}{b^{}}=1$ . . The values of $a^{}$ and $b^{}$ are9 and25, respectively.
3. The center of the ellipse is at the origin (0,0) because there are no shifts in the x or y direction in the equation.

STEP 2

The semi-major axis and semi-minor axis are given by the square root of $a^{2}$ and $b^{2}$ , respectively. $a = \sqrt{a^{2}}$ $b = \sqrt{b^{2}}$

STEP 3

Substitute the given values of $a^{2}$ and $b^{2}$ into the equations.
$a = \sqrt{9}$ $b = \sqrt{25}$

STEP 4

Calculate the values of $a$ and $b$ .
$a = \sqrt{9} =3$ $b = \sqrt{25} =$

STEP 5

The semi-major axis is the larger of the two values $a$ and $b$ . In this case, $b$ is larger, so the semi-major axis is5 and the semi-minor axis is3.

STEP 6

The foci of the ellipse are given by the equation $c = \sqrt{b^{2}-a^{2}}$ where $c$ is the distance from the center to each focus.

STEP 7

Substitute the values of $a$ and $b$ into the equation for $c$ .
$c = \sqrt{5^{2}-3^{2}}$

STEP 8

Calculate the value of $c$ .
$c = \sqrt{25-} = \sqrt{16} =4$

STEP 9

The foci are located at $(, \pm c)$ because the major axis is vertical. Substitute the value of $c$ into the coordinates.
$oci = (, \pm4)$

STEP 10

The eccentricity of the ellipse is given by the equation $e = \frac{c}{b}$ .

STEP 11

Substitute the values of $c$ and $b$ into the equation for $e$ .
$e = \frac{4}{5}$

STEP 12

The eccentricity of the ellipse is $e = \frac{4}{5}$ .
So, the properties of the ellipse are. Semi-major axis52. Semi-minor axis. Foci (0,4) and (0, -4)
4. Eccentricity $\frac{4}{5}$

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QuestionDetermine the properties of the ellipse defined by $\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$ .

Studdy Solution

STEP 1

Assumptions1. The given equation is of an ellipse in standard form, $\frac{x^{}}{a^{}}+\frac{y^{}}{b^{}}=1$ . . The values of $a^{}$ and $b^{}$ are9 and25, respectively.
3. The center of the ellipse is at the origin (0,0) because there are no shifts in the x or y direction in the equation.

STEP 2

The semi-major axis and semi-minor axis are given by the square root of $a^{2}$ and $b^{2}$ , respectively. $a = \sqrt{a^{2}}$ $b = \sqrt{b^{2}}$

STEP 3

Substitute the given values of $a^{2}$ and $b^{2}$ into the equations.
$a = \sqrt{9}$ $b = \sqrt{25}$

STEP 4

Calculate the values of $a$ and $b$ .
$a = \sqrt{9} =3$ $b = \sqrt{25} =$

STEP 5

The semi-major axis is the larger of the two values $a$ and $b$ . In this case, $b$ is larger, so the semi-major axis is5 and the semi-minor axis is3.

STEP 6

The foci of the ellipse are given by the equation $c = \sqrt{b^{2}-a^{2}}$ where $c$ is the distance from the center to each focus.

STEP 7

Substitute the values of $a$ and $b$ into the equation for $c$ .
$c = \sqrt{5^{2}-3^{2}}$

STEP 8

Calculate the value of $c$ .
$c = \sqrt{25-} = \sqrt{16} =4$

STEP 9

The foci are located at $(, \pm c)$ because the major axis is vertical. Substitute the value of $c$ into the coordinates.
$oci = (, \pm4)$

STEP 10

The eccentricity of the ellipse is given by the equation $e = \frac{c}{b}$ .

STEP 11

Substitute the values of $c$ and $b$ into the equation for $e$ .
$e = \frac{4}{5}$

STEP 12

The eccentricity of the ellipse is $e = \frac{4}{5}$ .
So, the properties of the ellipse are. Semi-major axis52. Semi-minor axis. Foci (0,4) and (0, -4)
4. Eccentricity $\frac{4}{5}$

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QuestionDetermine the properties of the ellipse defined by x29+y225=1\frac{x^{2}}{9}+\frac{y^{2}}{25}=19x2​+25y2​=1.

Studdy Solution

STEP 1

STEP 2

The semi-major axis and semi-minor axis are given by the square root of a2a^{2}a2 and b2b^{2}b2, respectively.a=a2a = \sqrt{a^{2}}a=a2​b=b2b = \sqrt{b^{2}}b=b2​

STEP 3

Substitute the given values of a2a^{2}a2 and b2b^{2}b2 into the equations.a=9a = \sqrt{9}a=9​b=25b = \sqrt{25}b=25​

STEP 4

Calculate the values of aaa and bbb.a=9=3a = \sqrt{9} =3a=9​=3b=25=b = \sqrt{25} =b=25​=

STEP 5

The semi-major axis is the larger of the two values aaa and bbb. In this case, bbb is larger, so the semi-major axis is5 and the semi-minor axis is3.

STEP 6

The foci of the ellipse are given by the equation c=b2−a2c = \sqrt{b^{2}-a^{2}}c=b2−a2​ where ccc is the distance from the center to each focus.

STEP 7

Substitute the values of aaa and bbb into the equation for ccc.c=52−32c = \sqrt{5^{2}-3^{2}}c=52−32​

STEP 8

Calculate the value of ccc.c=25−=16=4c = \sqrt{25-} = \sqrt{16} =4c=25−​=16​=4

STEP 9

The foci are located at (,±c)(, \pm c)(,±c) because the major axis is vertical. Substitute the value of ccc into the coordinates.oci=(,±4)oci = (, \pm4)oci=(,±4)

STEP 10

The eccentricity of the ellipse is given by the equation e=cbe = \frac{c}{b}e=bc​.

STEP 11

Substitute the values of ccc and bbb into the equation for eee.e=45e = \frac{4}{5}e=54​

STEP 12

The eccentricity of the ellipse is e=45e = \frac{4}{5}e=54​.So, the properties of the ellipse are. Semi-major axis52. Semi-minor axis. Foci (0,4) and (0, -4)4. Eccentricity 45\frac{4}{5}54​

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QuestionDetermine the properties of the ellipse defined by $\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$ .

The semi-major axis and semi-minor axis are given by the square root of $a^{2}$ and $b^{2}$ , respectively. $a = \sqrt{a^{2}}$ $b = \sqrt{b^{2}}$

Substitute the given values of $a^{2}$ and $b^{2}$ into the equations.
$a = \sqrt{9}$ $b = \sqrt{25}$

Calculate the values of $a$ and $b$ .
$a = \sqrt{9} =3$ $b = \sqrt{25} =$

The semi-major axis is the larger of the two values $a$ and $b$ . In this case, $b$ is larger, so the semi-major axis is5 and the semi-minor axis is3.

The foci of the ellipse are given by the equation $c = \sqrt{b^{2}-a^{2}}$ where $c$ is the distance from the center to each focus.

Substitute the values of $a$ and $b$ into the equation for $c$ .
$c = \sqrt{5^{2}-3^{2}}$

Calculate the value of $c$ .
$c = \sqrt{25-} = \sqrt{16} =4$

The foci are located at $(, \pm c)$ because the major axis is vertical. Substitute the value of $c$ into the coordinates.
$oci = (, \pm4)$

The eccentricity of the ellipse is given by the equation $e = \frac{c}{b}$ .

Substitute the values of $c$ and $b$ into the equation for $e$ .
$e = \frac{4}{5}$

The eccentricity of the ellipse is $e = \frac{4}{5}$ .
So, the properties of the ellipse are. Semi-major axis52. Semi-minor axis. Foci (0,4) and (0, -4)
4. Eccentricity $\frac{4}{5}$