How do i graph this hyperbola?
X^2+6x-3y^2=0
complete the square
x^2 + 6x + 9 - 3y^2 = 9
(x+3)^2 - 3y^2 = 9
divide by 9
(x+3)^2 /9 - y^2/3 = 1
centre (-3,0)
a = 3, b = √3
vertices at (3,0) and (-3,0)
faintly sketch a rectangle with corners
at (±3,±√3)
draw in the diagonals and extend them
sketch the hyperbola with the diagonals as asymptotes and the vertices as noted above
except the whole thing is shifted 3 units to the left - because it's (x+3)^2
To graph a hyperbola, we need to follow a few steps:
Step 1: Rewrite the equation in the standard form.
Step 2: Identify the center and the vertices.
Step 3: Determine the foci and the asymptotes.
Step 4: Plot the points and draw the hyperbola.
Let's go through each step:
Step 1: Rewrite the equation in standard form.
The standard form of a hyperbola equation is (x-h)^2/a^2 - (y-k)^2/b^2 = 1 or (y-k)^2/b^2 - (x-h)^2/a^2 = 1, where (h, k) represents the center of the hyperbola.
For the given equation x^2 + 6x - 3y^2 = 0, we need to complete the square for both the x and y terms.
Rearrange the equation:
x^2 + 6x = 3y^2
We need to add and subtract appropriate terms inside the parentheses to make it a perfect square trinomial:
x^2 + 6x + 9 = 3y^2 + 9
(x + 3)^2 = 3(y^2 + 3)
Divide both sides by 3:
(x + 3)^2/3 = y^2 + 3
Now, rewrite the equation in standard form by dividing both the x and y terms by the same value, so that the right side equals 1:
(x + 3)^2/9 - (y^2 + 3)/3 = 1
Step 2: Identify the center and the vertices.
Comparing the equation with the standard form, we can identify the center as (-3, 0), because (h, k) is the opposite sign of the values inside the parentheses.
The distance from the center to the vertices is given by "a" in the standard form. In this case, a = √9 = 3.
So, the center of the hyperbola is (-3, 0), and the vertices are (-3 - 3, 0) and (-3 + 3, 0), which simplifies to (-6, 0) and (0, 0) respectively.
Step 3: Determine the foci and the asymptotes.
To find the foci, we need to calculate "c" in the standard form, where c = √(a^2 + b^2).
Here, a = 3, and to find b, we rearrange the standard form equation: (x + 3)^2/9 - (y^2 + 3)/3 = 1
Multiply through by 9 to eliminate the fractions:
(x + 3)^2 - 3(y^2 + 3) = 9
(x + 3)^2 - 3y^2 - 9 = 9
(x + 3)^2 - 3y^2 = 18
Comparing this with the standard form equation for a hyperbola, we find that b = √3.
Calculating c:
c = √(3^2 + √3^2) = √(9 + 3) = √12 = 2√3
The foci are located at the same x-coordinate as the center but differ in the y-coordinate by a distance of 2√3. So, the foci are at (-3, -2√3) and (-3, 2√3).
To determine the asymptotes of the hyperbola, we use the formula: y = ±(b/a)(x-h) + k.
Plugging in the values:
y = ± (√3/3)(x + 3) + 0
Simplifying:
y = ± (√3/3)x - √3
Step 4: Plot the points and draw the hyperbola.
Now that we have all the necessary points and information, we can plot them on a graph and sketch the hyperbola.
- Plot the center at (-3, 0).
- Draw the vertices at (-6, 0) and (0, 0).
- Mark the foci at (-3, -2√3) and (-3, 2√3).
- Draw the asymptotes using the equations: y = (√3/3)x - √3 and y = -(√3/3)x - √3.
- Sketch the hyperbola using the vertices, foci, and asymptotes as guides.
Remember, a hyperbola consists of two separate branches that open up and down from the center. Draw smooth curves that pass through the vertices, approach the asymptotes, and get closer to the foci as they flow away from the center.
And there you have it! You have successfully graphed the hyperbola x^2 + 6x - 3y^2 = 0.