Find the Foci 4x^2-y^2-24x-6y+23=0
The question provided relates to the field of algebra and geometry, and specifically to the study of conic sections. It asks to find the coordinates of the focal points (foci) of a conic section, which in this case is defined by the given quadratic equation 4x^2 - y^2 - 24x - 6y + 23 = 0. This equation represents a hyperbola, and the task is to determine the two points, called foci, which are characteristic to this type of conic section. Identifying the foci involves completing the square for both the x and y terms, bringing the equation to a standard form that clearly identifies its center, axes, and the relationship between the semi-major axis and the distance to the foci, so that the positions of the foci can be accurately calculated.
$4 x^{2} - y^{2} - 24 x - 6 y + 23 = 0$
Answer
Solution:
Step:1
Convert the given equation to the standard form of a hyperbola.
Step:1.1
Isolate the constant term by subtracting $23$ from both sides: $4x^2 - y^2 - 24x - 6y = -23$.
Step:1.2
Complete the square for the $x$-terms, $4x^2 - 24x$.
Step:1.2.1
Identify coefficients: $a = 4$, $b = -24$, $c = 0$.
Step:1.2.2
Use the vertex form $a(x + d)^2 + e$ for transformation.
Step:1.2.3
Calculate $d$ with $d = \frac{b}{2a}$.
Step:1.2.3.1
Insert $a$ and $b$ into the formula: $d = \frac{-24}{2 \cdot 4}$.
Step:1.2.3.2
Simplify the fraction.
Step:1.2.3.2.1
Factor out $2$: $d = \frac{2 \cdot -12}{2 \cdot 4}$.
Step:1.2.3.2.1.2
Reduce common factors: $d = \frac{-12}{4}$.
Step:1.2.3.2.2
Divide: $d = -3$.
Step:1.2.4
Determine $e$ using $e = c - \frac{b^2}{4a}$.
Step:1.2.4.1
Plug in $c$, $b$, and $a$: $e = 0 - \frac{(-24)^2}{4 \cdot 4}$.
Step:1.2.4.2
Perform the operations.
Step:1.2.4.2.1
Calculate each term: $e = 0 - \frac{576}{16}$.
Step:1.2.4.2.2
Subtract: $e = -36$.
Step:1.2.5
Insert $a$, $d$, and $e$ into the vertex form: $4(x - 3)^2 - 36$.
Step:1.3
Replace $4x^2 - 24x$ in the original equation with the completed square: $4(x - 3)^2 - 36 - y^2 - 6y = -23$.
Step:1.4
Add $36$ to both sides to balance the equation: $4(x - 3)^2 - y^2 - 6y = 13$.
Step:1.5
Complete the square for the $y$-terms, $-y^2 - 6y$.
Step:1.5.1
Identify coefficients: $a = -1$, $b = -6$, $c = 0$.
Step:1.5.2
Use the vertex form $a(y + d)^2 + e$ for transformation.
Step:1.5.3
Calculate $d$ with $d = \frac{b}{2a}$.
Step:1.5.3.1
Insert $a$ and $b$ into the formula: $d = \frac{-6}{2 \cdot -1}$.
Step:1.5.3.2
Simplify the fraction: $d = 3$.
Step:1.5.4
Determine $e$ using $e = c - \frac{b^2}{4a}$.
Step:1.5.4.1
Plug in $c$, $b$, and $a$: $e = 0 - \frac{(-6)^2}{4 \cdot -1}$.
Step:1.5.4.2
Perform the operations: $e = 9$.
Step:1.5.5
Insert $a$, $d$, and $e$ into the vertex form: $-(y + 3)^2 + 9$.
Step:1.6
Replace $-y^2 - 6y$ in the original equation with the completed square: $4(x - 3)^2 - (y + 3)^2 + 9 = 13$.
Step:1.7
Subtract $9$ from both sides to balance the equation: $4(x - 3)^2 - (y + 3)^2 = 4$.
Step:1.8
Simplify the right side of the equation.
Step:1.9
Divide each term by $4$ to get the equation in standard form: $(x - 3)^2 - \frac{(y + 3)^2}{4} = 1$.
Step:1.10
The standard form of the hyperbola is now achieved.
Step:2
Identify the standard form of a hyperbola: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$.
Step:3
Match the equation's components with the standard form to find $h$, $k$, $a$, and $b$: $h = 3$, $k = -3$, $a = 1$, $b = 2$.
Step:4
Calculate the distance from the center to a focus, $c$.
Step:4.1
Use the formula $c = \sqrt{a^2 + b^2}$.
Step:4.2
Substitute the values for $a$ and $b$: $c = \sqrt{1^2 + 2^2}$.
Step:4.3
Simplify the expression: $c = \sqrt{5}$.
Step:5
Determine the coordinates of the foci.
Step:5.1
The first focus is found by adding $c$ to $h$: $(h + c, k)$.
Step:5.2
Insert the values and simplify: $(3 + \sqrt{5}, -3)$.
Step:5.3
The second focus is found by subtracting $c$ from $h$: $(h - c, k)$.
Step:5.4
Insert the values and simplify: $(3 - \sqrt{5}, -3)$.
Step:5.5
The foci of the hyperbola are $(3 + \sqrt{5}, -3)$ and $(3 - \sqrt{5}, -3)$.
Step:6
The process is complete, and the foci of the hyperbola have been found.
Knowledge Notes:
A hyperbola is a type of conic section that appears when a plane intersects both nappes (the upper and lower portions) of a cone.
The standard form of a hyperbola with a horizontal transverse axis is $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$, and with a vertical transverse axis is $\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$, where $(h, k)$ is the center of the hyperbola.
Completing the square is a method used to transform a quadratic equation into a perfect square trinomial plus a constant.
The distance from the center to a focus of a hyperbola is given by $c = \sqrt{a^2 + b^2}$, where $a$ is the distance from the center to a vertex and $b$ is the distance from the center to a co-vertex.
The foci of a hyperbola are located along the major axis, symmetrically spaced from the center, at a distance of $c$ from the center.
