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Write the standard form of the equation of the hyperbola [tex]$4x^2 - 9y^2 - 16x - 36y - 56 = 0$[/tex].

Place the signs and values in the correct places in the standard form of the equation.
4
3
2
9
[tex]+[/tex]
[tex]-[/tex]


Sagot :

To convert the given hyperbola equation [tex]\(4x^2 - 9y^2 - 16x - 36y - 56 = 0\)[/tex] into its standard form, we need to complete the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms. Here's a step-by-step process:

1. Group [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together:
[tex]\[ 4x^2 - 16x - 9y^2 - 36y - 56 = 0 \][/tex]

2. Complete the square for [tex]\(x\)[/tex] terms:
- Factor out the coefficient of [tex]\(x^2\)[/tex], which is 4:
[tex]\[ 4(x^2 - 4x) \][/tex]
- To complete the square inside the parentheses, take half of the coefficient of [tex]\(x\)[/tex] (which is -4), square it to get 4, and then add and subtract 4 inside the parentheses:
[tex]\[ 4(x^2 - 4x + 4 - 4) = 4((x-2)^2 - 4) = 4(x-2)^2 - 16 \][/tex]

3. Complete the square for [tex]\(y\)[/tex] terms:
- Factor out the coefficient of [tex]\(y^2\)[/tex], which is -9:
[tex]\[ -9(y^2 + 4y) \][/tex]
- To complete the square inside the parentheses, take half of the coefficient of [tex]\(y\)[/tex] (which is 4), square it to get 4, and then add and subtract 4 inside the parentheses:
[tex]\[ -9(y^2 + 4y + 4 - 4) = -9((y + 2)^2 - 4) = -9(y + 2)^2 + 36 \][/tex]

4. Substitute the completed squares back into the equation:
[tex]\[ 4(x - 2)^2 - 16 - 9(y + 2)^2 + 36 - 56 = 0 \][/tex]

5. Combine the constant terms (i.e., [tex]\(-16\)[/tex], [tex]\(36\)[/tex], and [tex]\(-56\)[/tex]):
[tex]\[ 4(x - 2)^2 - 9(y + 2)^2 - 16 + 36 - 56 = 0 \implies 4(x - 2)^2 - 9(y + 2)^2 - 36 = 0 \][/tex]

6. Move the constant to the other side to write the equation in standard form:
[tex]\[ 4(x - 2)^2 - 9(y + 2)^2 - 16 = 1 \][/tex]

Now we have successfully converted the given hyperbola equation into its standard form:
[tex]\[ 4(x - 2)^2 - 9(y + 2)^2 - 16 = 1 \][/tex]

Therefore, the final standard form of the equation is:
[tex]\[ 4(x - 2)^2 \, - \, 9(y + 2)^2 \, - \, 16 \, = \, 1 \][/tex]