Find the Foci (x^2)/4+(y^2)/5=1

The problem asks to determine the coordinates of the foci for the given ellipse, which is represented by its equation in the standard form. The standard form of an ellipse's equation along the Cartesian plane is given by (x^2)/a^2 + (y^2)/b^2 = 1, where the lengths of the semi-major axis and semi-minor axis are represented by 'a' and 'b' respectively. Depending on the values of 'a' and 'b', the ellipse can be oriented along the x-axis or y-axis. The foci of an ellipse are two points located along the major axis of the ellipse, equidistant from the ellipse's center, and the distance of each focus from the center is given by the square root of the absolute difference of the squares of 'a' and 'b' (c = sqrt(|a^2 - b^2|)). The question requires finding the values of these points using the given equation parameters.

$\frac{x^{2}}{4} + \frac{y^{2}}{5} = 1$

Answer

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Solution:

Step:1

Normalize the equation to have 1 on the right side, conforming to the standard form of an ellipse or hyperbola equation. The given equation is already in the desired form: $\frac{x^{2}}{4} + \frac{y^{2}}{5} = 1$ .

Step:2

Recognize that the equation represents an ellipse. The general equation for an ellipse is $\frac{(x - h)^{2}}{b^{2}} + \frac{(y - k)^{2}}{a^{2}} = 1$ , where $(h, k)$ is the center, and $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively.

Step:3

Identify the values corresponding to $a$ , $b$ , $h$ , and $k$ from the given equation, where $a$ is the semi-major axis, $b$ is the semi-minor axis, and $(h, k)$ is the center of the ellipse. We find $a = \sqrt{5}$ , $b = 2$ , $h = 0$ , and $k = 0$ .

Step:4

Determine the distance $c$ from the center to a focus of the ellipse.

Step:4.1

Use the formula $c = \sqrt{a^{2} - b^{2}}$ to calculate the distance.

Step:4.2

Plug in the values for $a$ and $b$ : $c = \sqrt{(\sqrt{5})^{2} - 2^{2}}$ .

Step:4.3

Simplify the expression.

Step:4.3.1

Convert $(\sqrt{5})^{2}$ to $5$ .

Step:4.3.1.1

Express $\sqrt{5}$ as $5^{\frac{1}{2}}$ and apply the formula: $c = \sqrt{(5^{\frac{1}{2}})^{2} - 2^{2}}$ .

Step:4.3.1.2

Use the power rule to simplify: $c = \sqrt{5^{\frac{1}{2} \cdot 2} - 2^{2}}$ .

Step:4.3.1.3

Combine the exponents: $c = \sqrt{5^{1} - 2^{2}}$ .

Step:4.3.1.4

Simplify the expression: $c = \sqrt{5 - 2^{2}}$ .

Step:4.3.2

Finalize the simplification.

Step:4.3.2.1

Square the number 2: $c = \sqrt{5 - 4}$ .

Step:4.3.2.2

Subtract 4 from 5: $c = \sqrt{1}$ .

Step:4.3.2.3

The square root of 1 is 1: $c = 1$ .

Step:5

Locate the foci of the ellipse.

Step:5.1

To find the first focus, add $c$ to $k$ : $(h, k + c)$ .

Step:5.2

Insert the known values for $h$ , $k$ , and $c$ : $(0, 0 + 1)$ .

Step:5.3

The result simplifies to $(0, 1)$ .

Step:5.4

To find the second focus, subtract $c$ from $k$ : $(h, k - c)$ .

Step:5.5

Insert the known values for $h$ , $k$ , and $c$ : $(0, 0 - 1)$ .

Step:5.6

The result simplifies to $(0, - 1)$ .

Step:5.7

An ellipse has two foci: $(Focus)_{1} : (0, 1)$ and $(Focus)_{2} : (0, - 1)$ .

Step:6

The foci of the ellipse are at $(0, 1)$ and $(0, - 1)$ .

Knowledge Notes:

Ellipse Equation: The standard form of an ellipse 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.
Foci of an Ellipse: The foci of an ellipse are two fixed points located along the major axis, equidistant from the center. The distance from the center to each focus is given by $c = \sqrt{a^{2} - b^{2}}$ .
Major and Minor Axes: In an ellipse, the major axis is the longest diameter, and the minor axis is the shortest diameter. The lengths of the semi-major axis and the semi-minor axis are denoted by $a$ and $b$ , respectively.
Simplifying Square Roots: When simplifying expressions involving square roots, remember that $\sqrt{a^{2}} = a$ and $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ .
Finding Foci Coordinates: The coordinates of the foci of an ellipse can be found by adding and subtracting the value of $c$ from the $y$ -coordinate of the center if the major axis is vertical, or from the $x$ -coordinate if the major axis is horizontal. In this case, since $h$ and $k$ are both zero, the foci are at $(0, k + c)$ and $(0, k - c)$ .