Graph the conic section of 9x^2-36y^2=324
To graph the conic section 9x^2-36y^2=324, we can start by rearranging it into a standard form:
9x^2/324 - 36y^2/324 = 1
x^2/36 - y^2/9 = 1
This is the equation of a hyperbola with a horizontal transverse axis.
To graph it, we can plot the center (0,0), which is the midpoint of the two branches of the hyperbola. Then, we can locate the vertices, which are a distance of 6 units to the right and left of the center, giving us the points (6,0) and (-6,0). We can also locate the foci, which are a distance of sqrt(36+9) = 3sqrt(5) units to the right and left of the center, giving us the points (3sqrt(5),0) and (-3sqrt(5),0).
Finally, we can sketch the two branches of the hyperbola, which are asymptotic to the lines y = 3x/2 and y = -3x/2 as x becomes infinitely large in either direction. The resulting graph should look like this:
To graph the conic section given by the equation 9x^2 - 36y^2 = 324, we can start by rearranging the equation to get it in standard form:
9x^2 - 36y^2 = 324
Divide both sides by 324 to simplify the equation:
x^2/36 - y^2/9 = 1
Now, let's analyze the equation to determine the conic section. Since the x^2 term has a positive coefficient and the y^2 term has a negative coefficient, we can conclude that it is a hyperbola.
The standard form of a hyperbola can be written as:
(x - h)^2/a^2 - (y - k)^2/b^2 = 1
Comparing this with the given equation, we can see that:
h = 0, k = 0, a = 6, and b = 3
Now, let's plot the points on a graph:
1. The center of the hyperbola is at the point (h, k) = (0, 0). Mark this point on the graph.
2. The distance from the center to the vertices along the x-axis is a = 6. So, measure 6 units to the left and right of the center point along the x-axis and mark these points as the vertices V1 and V2.
3. The distance from the center to the foci along the x-axis can be found using the formula c^2 = a^2 + b^2, where c is the distance from the center to the foci. Substituting the given values, we can find that c^2 = 36 + 9 = 45. So, c = √45 ≈ 6.71.
4. Measure 6.71 units to the left and right of the center point along the x-axis and mark these points as the foci F1 and F2.
5. Connect the points V1, V2, and the foci F1, F2 with dashed lines. This will be the transverse axis of the hyperbola.
6. Measure the distance from the center along the y-axis, which is b = 3. Mark the points above and below the center along the y-axis as the co-vertices C1 and C2.
7. Draw a dashed line connecting the co-vertices C1 and C2. This will be the conjugate axis of the hyperbola.
8. Finally, sketch the hyperbola using the vertices, foci, and the dashed lines representing the transverse and conjugate axes.
The graph of the conic section 9x^2 - 36y^2 = 324 will be a hyperbola whose shape depends on the values of a and b.