Math / Algebra

QuestionConvert the following equation for a hyperbola into standard form. $16 x^{2}-96 x-y^{2}-4 y+124=0$
Select the correct answer below: $\frac{(y-2)^{2}}{16}-(x+3)^{2}=1$ $(x-3)^{2}-\frac{(y+2)^{2}}{16}=1$ $\frac{(y-2)^{2}}{16}-(x-3)^{2}=1$ $(x+3)^{2}-\frac{(y-2)^{2}}{16}=1$ $(x-3)^{2}-\frac{(y-2)^{2}}{16}=1$ $\frac{(y+2)^{2}}{16}-(x-3)^{2}=1$

Studdy Solution

STEP 1

What is this asking? We need to rearrange this hyperbola equation into its standard form, which helps us understand its shape and key features! Watch out! Completing the square can be tricky, so double-check those calculations!
Also, remember the standard form of a hyperbola equation.

STEP 2

1. Group Terms
2. Complete the Square
3. Standard Form

STEP 3

Let's group our $x$ terms and $y$ terms together.
This sets us up perfectly for completing the square in the next step! $16x^2 - 96x - y^2 - 4y + 124 = 0$ becomes $(16x^2 - 96x) + (-y^2 - 4y) + 124 = 0$ .

STEP 4

For the $x$ terms, we factor out the $\textbf{16}$ from $16x^2 - 96x$ , giving us $16(x^2 - 6x)$ .
To complete the square, take half of the coefficient of our $x$ term (which is $-6$ ), square it $( \frac{-6}{2} )^2 = (-3)^2 = 9$ , and add it inside the parenthesis.
Since we're adding $16 \cdot 9 = \textbf{144}$ to the left side, we also add $\textbf{144}$ to the right side to keep the equation balanced.

STEP 5

Now, our equation looks like this: $16(x^2 - 6x + 9) + (-y^2 - 4y) + 124 = 144$ . The $x$ part is now a perfect square: $16(x-3)^2$ .

STEP 6

For the $y$ terms, we factor out $\textbf{-1}$ from $-y^2 - 4y$ , giving us $-(y^2 + 4y)$ .
To complete the square, take half of the coefficient of our $y$ term (which is $4$ ), square it $(\frac{4}{2})^2 = 2^2 = 4$ , and add it inside the parenthesis.
Since we're adding $-1 \cdot 4 = \textbf{-4}$ to the left side, we also add $\textbf{-4}$ to the right side to keep the equation balanced.

STEP 7

Now, our equation looks like this: $16(x-3)^2 -(y^2 + 4y + 4) + 124 = 144 - 4$ which simplifies to $16(x-3)^2 -(y+2)^2 + 124 = 140$ . The $y$ part is now a perfect square: $-(y+2)^2$ .

STEP 8

Let's move that constant to the right side by subtracting $\textbf{124}$ from both sides: $16(x-3)^2 - (y+2)^2 = 140 - 124$ $16(x-3)^2 - (y+2)^2 = 16$

STEP 9

To get to standard form, we need a $1$ on the right side.
Let's divide both sides by $\textbf{16}$ : $\frac{16(x-3)^2}{16} - \frac{(y+2)^2}{16} = \frac{16}{16}$ $\frac{16}{16} \cdot (x-3)^2 - \frac{(y+2)^2}{16} = 1$ $(x-3)^2 - \frac{(y+2)^2}{16} = 1$

STEP 10

The standard form of the hyperbola equation is $(x-3)^2 - \frac{(y+2)^2}{16} = 1$ .

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Studdy Solution

STEP 1

What is this asking? We need to rearrange this hyperbola equation into its standard form, which helps us understand its shape and key features! Watch out! Completing the square can be tricky, so double-check those calculations!
Also, remember the standard form of a hyperbola equation.

STEP 2

1. Group Terms
2. Complete the Square
3. Standard Form

STEP 3

Let's group our $x$ terms and $y$ terms together.
This sets us up perfectly for completing the square in the next step! $16x^2 - 96x - y^2 - 4y + 124 = 0$ becomes $(16x^2 - 96x) + (-y^2 - 4y) + 124 = 0$ .

STEP 4

STEP 5

Now, our equation looks like this: $16(x^2 - 6x + 9) + (-y^2 - 4y) + 124 = 144$ . The $x$ part is now a perfect square: $16(x-3)^2$ .

STEP 6

STEP 7

Now, our equation looks like this: $16(x-3)^2 -(y^2 + 4y + 4) + 124 = 144 - 4$ which simplifies to $16(x-3)^2 -(y+2)^2 + 124 = 140$ . The $y$ part is now a perfect square: $-(y+2)^2$ .

STEP 8

Let's move that constant to the right side by subtracting $\textbf{124}$ from both sides: $16(x-3)^2 - (y+2)^2 = 140 - 124$ $16(x-3)^2 - (y+2)^2 = 16$

STEP 9

To get to standard form, we need a $1$ on the right side.
Let's divide both sides by $\textbf{16}$ : $\frac{16(x-3)^2}{16} - \frac{(y+2)^2}{16} = \frac{16}{16}$ $\frac{16}{16} \cdot (x-3)^2 - \frac{(y+2)^2}{16} = 1$ $(x-3)^2 - \frac{(y+2)^2}{16} = 1$

STEP 10

The standard form of the hyperbola equation is $(x-3)^2 - \frac{(y+2)^2}{16} = 1$ .

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Studdy Solution

STEP 1

What is this asking? We need to rearrange this hyperbola equation into its standard form, which helps us understand its shape and key features! Watch out! Completing the square can be tricky, so double-check those calculations!Also, remember the standard form of a hyperbola equation.

STEP 2

1. Group Terms2. Complete the Square3. Standard Form

STEP 3

STEP 4

STEP 5

Now, our equation looks like this: 16(x2−6x+9)+(−y2−4y)+124=14416(x^2 - 6x + 9) + (-y^2 - 4y) + 124 = 14416(x2−6x+9)+(−y2−4y)+124=144. The xxx part is now a perfect square: 16(x−3)216(x-3)^216(x−3)2.

STEP 6

STEP 7

STEP 8

Let's move that constant to the right side by subtracting 124\textbf{124}124 from both sides: 16(x−3)2−(y+2)2=140−12416(x-3)^2 - (y+2)^2 = 140 - 12416(x−3)2−(y+2)2=140−124 16(x−3)2−(y+2)2=1616(x-3)^2 - (y+2)^2 = 1616(x−3)2−(y+2)2=16

STEP 9

STEP 10

The standard form of the hyperbola equation is (x−3)2−(y+2)216=1(x-3)^2 - \frac{(y+2)^2}{16} = 1(x−3)2−16(y+2)2​=1.

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What is this asking? We need to rearrange this hyperbola equation into its standard form, which helps us understand its shape and key features! Watch out! Completing the square can be tricky, so double-check those calculations!
Also, remember the standard form of a hyperbola equation.

1. Group Terms
2. Complete the Square
3. Standard Form

Now, our equation looks like this: $16(x^2 - 6x + 9) + (-y^2 - 4y) + 124 = 144$ . The $x$ part is now a perfect square: $16(x-3)^2$ .

Let's move that constant to the right side by subtracting $\textbf{124}$ from both sides: $16(x-3)^2 - (y+2)^2 = 140 - 124$ $16(x-3)^2 - (y+2)^2 = 16$

The standard form of the hyperbola equation is $(x-3)^2 - \frac{(y+2)^2}{16} = 1$ .