Find the Foci 4x^2-y^2-24x-6y+23=0

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.