fin the center ,principal axis vertices, sketch the graph of 4x^-y^2 +56 x +2y+195=0
let's complete the square ...
4x^2 + 56x - (y^2 - 2y ) = -195
4(x^2 + 14x + 49) - (y^2 - 2y + 1) = -195 + 4(49) - 1
4(x+7)^2 - (y-1)^2 = 0
ahh, we have a degenerated hyperbola
the hyperbola has degenerated into its asymtotes
notice we have a difference of squares = 0
so
2(x+7) + (y-1) = 0 or 2(x+7) - (y-1) = 0
2x + 14 + y - 1 = 0 or 2x + 14 - y + 1 = 0
2x + y = -13 or 2x - y = -15
the "centre" would be the intersection of the two lines.
To find the center, principal axis, and vertices of the graph of the equation 4x^-y^2 + 56x + 2y + 195 = 0, you need to rewrite the equation in a more standard form called the equation of a conic section.
Step 1: Group the terms with x and y together
Rearrange the equation:
4x^2 + 56x - y^2 + 2y + 195 = 0
Step 2: Complete the square for x
To complete the square for x, divide the coefficient of x by 2 and square it. Then add it to both sides of the equation.
4(x^2 + 14x) - y^2 + 2y + 195 = 0
4(x^2 + 14x + 49) - y^2 + 2y + 195 - 4(49) = 0
4(x + 7)^2 - y^2 + 2y + 195 - 196 = 0
4(x + 7)^2 - y^2 + 2y - 1 = 0
Step 3: Complete the square for y
To complete the square for y, divide the coefficient of y by 2 and square it. Then add it to both sides of the equation.
4(x + 7)^2 + (2y + 1)^2 - 1 - (2y + 1)^2 = 0
4(x + 7)^2 + (2y + 1)^2 - (2y + 1)^2 - 1 = 0
4(x + 7)^2 + (2y + 1)^2 - (2y + 1)^2 - 1 = 0
4(x + 7)^2 + (2y + 1)^2 - (2y + 1)^2 - 1 = 0
Step 4: Simplify the equation
Combine like terms:
4(x + 7)^2 - (2y + 1)^2 - 1 = 0
Step 5: Identify the conic section
By looking at the equation, we can see that it represents a hyperbola since there is a difference between the two variables (x and y).
Step 6: Find the center
The center of the hyperbola is given by (-h, -k). In this case, the center is (-7, -1).
Step 7: Find the principal axis
The principal axis represents the line passing through the center of the hyperbola. In this case, the principal axis is a vertical line since the x term is squared and has a positive coefficient.
Step 8: Find the vertices
The vertices of the hyperbola can be found by adding or subtracting the value of a from the center (h, k), where "a" represents the distance from the center to the vertices along the principal axis.
In this equation, the coefficient of (x + 7)^2 is 4, so a = √4 = 2. The vertices are then (-7, -1 ± 2) = (-7, -3) and (-7, 1).
Step 9: Sketch the graph
Plot the center (-7, -1), the vertices (-7, -3) and (-7, 1), and draw the hyperbola passing through these points.
The graph of the hyperbola would look similar to:
```
|
-3 + - - - - - - - - - + - - - - - -
|
-1 + - o - - - X - - - + - - - - - -
|
1 + - - o - - - - - - + - - - - - -
|
3 + - - - - - - - - - + - - - - - -
|
+--------------------------------------------->
-15 -12 -9 -6 -3 0 3 6 9 12 15
```
Note: The positive vertex is marked with a ("o") and the center is marked with a ("X").