Final answer:
The given equation is a hyperbola. The vertices are (±2/3, 0), the foci are (±2√2, 0), and the asymptotes are given by the equations y = ±(2/3)x.The correct option is c) Asymptotes only.
Step-by-step explanation:
The given equation 4x^2 - 9y^2 = 36 is a hyperbola because it has both x^2 and y^2 terms with different signs.
To find the vertices, we need to isolate either the x^2 or y^2 term. In this case, let's isolate the x^2 term:
4x^2 = 9y^2 + 36
x^2 = (9y^2 + 36)/4
x^2/9 = (y^2 + 4)/4
(x^2/9) - (y^2 + 4)/4 = 1
Now we can identify the values needed to determine the vertices, foci, and asymptotes:
- The center of the hyperbola is at (0, 0).
- The vertices are located at (±a, 0) where a is the square root of the value inside the denominator of the x^2 term. In this case, a = √(4/9) = 2/3, so the vertices are (±2/3, 0).
- The foci are located at (±c, 0) where c is the distance from the center to each focus. The value of 'c' can be found using the formula c^2 = a^2 + b^2, where a and b are the coefficients of x^2 and y^2, respectively. In this case, c^2 = 9(4/9) + 4 = 4 + 4 = 8, so c = √8 = 2√2. Therefore, the foci are at (±2√2, 0).
- The equation for the asymptotes of the hyperbola is y = ±(b/a)x, where b is the square root of the value inside the numerator of the y^2 term. In this case, b = √4 = 2, so the asymptotes are given by the equations y = ±(2/3)x.
The correct option is c) Asymptotes only.