Question

Identify the type of the equation 4x^2-y^2-16x-4y-4=0 and then change it to standard form.

Answer 1

Group the x terms together and complete the square.
Group the y terms together and complete the square.
Complete these two steps first.

Answer 2

kk

Answer 3

4x^2-y^2-16x-4y-4 = 0
(4x^2 - 16x) - (y^2 + 4y) = 4
4(x^2 - 4x) - (y^2 + 4y) = 4
complete the square of: (x^2 - 4x)
complete the square of: (y^2 + 4y)

Answer 4

got it

Answer 5

This equation represents a hyperbola whose standard form looks like this:\[
\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1
\]Completing the square will help in putting the equation in standard form.

Answer 6

How do you complete the square

Answer 7

\[
\text{Complete the square:} \\
x^2 - 4x = x^2 - 4x + 4 - 4 = (x-2)^2 - 4 \\
y^2 + 4y = y^2 + 4y + 4 - 4 = (y+2)^2 - 4 \\
\text{Substitute the above two in the earlier equation: } 4(x^2 - 4x) - (y^2 + 4y) = 4 \\
4{(x-2)^2 - 4} - {(y+2)^2 - 4} = 4 \\
4(x-2)^2 - 16 - (y+2)^2 + 4 = 4 \\
4(x-2)^2 - (y+2)^2 = 4 + 16 - 4 \\
4(x-2)^2 - (y+2)^2 = 16 \\
\text{Divide by 16 to make the right hand side 1:} \\
\frac{(x-2)^2}{4} - \frac{(y+2)^2}{16} = 1 \\
\text{That is the standard form of a hyperbola.}
\]