Find the Foci 49x^2+25y^2+490x-350y+1225=0

The problem is asking for the identification and calculation of the focal points (foci) of an ellipse. An ellipse is a geometric shape that looks like a stretched circle. The equation given is a general quadratic equation in two variables, x and y, which represents an ellipse in standard form after completing the square and dividing by the necessary constant. The foci of an ellipse are two points located along the major axis of the ellipse, equidistant from the center, and are central to the definition of an ellipse because the sum of distances from the foci to any point on the ellipse is constant. The question requires algebraic manipulation to rewrite the equation in standard form and then use the properties of an ellipse to find the coordinates of these focal points.

$49 x^{2} + 25 y^{2} + 490 x - 350 y + 1225 = 0$

Answer

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Solution:

Step:1

Transform the equation into the standard form of an ellipse.

Step:1.1

Remove $1225$ from both sides of the equation to get $49 x^{2} + 25 y^{2} + 490 x - 350 y = - 1225$ .

Step:1.2

Perform the square completion for $49 x^{2} + 490 x$ .

Step:1.2.1

Identify $a$ , $b$ , and $c$ from $a x^{2} + b x + c$ where $a = 49$ , $b = 490$ , and $c = 0$ .

Step:1.2.2

Refer to the standard form of a squared binomial $a (x + d)^{2} + e$ .

Step:1.2.3

Calculate $d$ using $d = \frac{b}{2 a}$ .

Step:1.2.3.1

Insert $a$ and $b$ into $d = \frac{b}{2 a}$ to obtain $d = \frac{490}{2 \cdot 49}$ .

Step:1.2.3.2

Simplify the fraction.

Step:1.2.3.2.1

Extract $2$ from $490$ to get $d = \frac{2 \cdot 245}{2 \cdot 49}$ .

Step:1.2.3.2.1.2

Eliminate the common factor to find $d = \frac{245}{49}$ .

Step:1.2.3.2.2

Reduce the fraction by removing the common factor between $245$ and $49$ .

Step:1.2.3.2.2.1

Isolate $49$ from $245$ to get $d = \frac{49 \cdot 5}{49}$ .

Step:1.2.3.2.2.2

Eliminate the common factor to conclude $d = 5$ .

Step:1.2.4

Determine $e$ using $e = c - \frac{b^{2}}{4 a}$ .

Step:1.2.4.1

Use the values of $c$ , $b$ , and $a$ in $e = c - \frac{b^{2}}{4 a}$ to get $e = 0 - \frac{490^{2}}{4 \cdot 49}$ .

Step:1.2.4.2

Simplify the expression to find $e = - 1225$ .

Step:1.2.5

Insert $a$ , $d$ , and $e$ into the vertex form to get $49 (x + 5)^{2} - 1225$ .

Step:1.3

Replace $49 x^{2} + 490 x$ with $49 (x + 5)^{2} - 1225$ in the original equation.

Step:1.4

Add $1225$ to both sides of the equation to isolate the square terms.

Step:1.5

Complete the square for $25 y^{2} - 350 y$ .

Step:1.5.1

From $a x^{2} + b x + c$ , find $a = 25$ , $b = - 350$ , and $c = 0$ .

Step:1.5.2

Refer to the squared binomial form $a (x + d)^{2} + e$ .

Step:1.5.3

Calculate $d$ with $d = \frac{b}{2 a}$ .

Step:1.5.3.1

Insert $a$ and $b$ into the formula to get $d = \frac{- 350}{2 \cdot 25}$ .

Step:1.5.3.2

Simplify to find $d = - 7$ .

Step:1.5.4

Determine $e$ using $e = c - \frac{b^{2}}{4 a}$ .

Step:1.5.4.1

Use $c$ , $b$ , and $a$ in the formula to get $e = 0 - \frac{(- 350)^{2}}{4 \cdot 25}$ .

Step:1.5.4.2

Simplify to conclude $e = - 1225$ .

Step:1.5.5

Insert $a$ , $d$ , and $e$ into the vertex form to obtain $25 (y - 7)^{2} - 1225$ .

Step:1.6

Replace $25 y^{2} - 350 y$ with $25 (y - 7)^{2} - 1225$ in the equation.

Step:1.7

Add $1225$ to both sides to balance the equation.

Step:1.8

Combine like terms on the right side to simplify.

Step:1.9

Divide each term by $1225$ to set the right side to $1$ .

Step:1.10

Simplify each term to achieve the standard form $\frac{(x + 5)^{2}}{25} + \frac{(y - 7)^{2}}{49} = 1$ .

Step:2

Recognize the equation as an ellipse and use this form to find the center, major axis, and minor axis.

Step:3

Align the values from the ellipse equation with the standard form to identify $a = 7$ , $b = 5$ , $h = - 5$ , and $k = 7$ .

Step:4

Calculate $c$ , the distance from the center to a focus.

Step:4.1

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

Step:4.2

Substitute $a$ and $b$ into the formula to get $\sqrt{7^{2} - 5^{2}}$ .

Step:4.3

Simplify to find $c = 2 \sqrt{6}$ .

Step:5

Determine the foci of the ellipse.

Step:5.1

Find the first focus by adding $c$ to $k$ , resulting in $(h, k + c)$ .

Step:5.2

Insert $h$ , $c$ , and $k$ to get the first focus at $(- 5, 7 + 2 \sqrt{6})$ .

Step:5.3

Find the second focus by subtracting $c$ from $k$ , resulting in $(h, k - c)$ .

Step:5.4

Use $h$ , $c$ , and $k$ to find the second focus at $(- 5, 7 - 2 \sqrt{6})$ .

Step:5.5

Simplify to finalize the foci coordinates.

Step:5.6

List both foci of the ellipse as ${Focus}_{1} : (- 5, 7 + 2 \sqrt{6})$ and ${Focus}_{2} : (- 5, 7 - 2 \sqrt{6})$ .

Step:6

The foci of the ellipse are at $(- 5, 7 + 2 \sqrt{6})$ and $(- 5, 7 - 2 \sqrt{6})$ .

Knowledge Notes:

Standard Form of an Ellipse: The standard form of an ellipse is given by $\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1$ where $(h, k)$ is the center of the ellipse, $a$ is the semi-major axis length, and $b$ is the semi-minor axis length.
Completing the Square: This technique is used to transform a quadratic expression into a perfect square trinomial. It involves adding and subtracting a particular value to complete the square.
Ellipse Foci: The foci of an ellipse are two fixed points inside the ellipse such that the sum of the distances from any point on the ellipse to the foci is constant. The distance from the center to a focus is given by $c = \sqrt{a^{2} - b^{2}}$ for an ellipse with the standard form above.
Vertex Form of a Parabola: The vertex form of a parabola is given by $a (x - h)^{2} + k$ where $(h, k)$ is the vertex of the parabola.
Simplifying Expressions: When simplifying expressions, common factors can be canceled, and operations such as addition, subtraction, multiplication, and division are performed to reduce the expression to its simplest form.
Radicals: When simplifying radicals, look for perfect square factors to simplify the square root. For example, $\sqrt{24}$ can be simplified to $2 \sqrt{6}$ because $24 = 4 \cdot 6$ and $\sqrt{4} = 2$ .