Final answer:
In solving the equation 16x² - 12y² = 192, we first normalize it to find the vertices at (3,0) and (-3,0). The foci are found using the formula c² = a² + b². The asymptotes are given by the lines y = ±(4/3.464)x.
Step-by-step explanation:
The student has been asked to find the vertices, foci, and asymptotes of the hyperbola represented by the equation 16x² - 12y² = 192. This is typically covered in high school algebra or pre-calculus courses. To solve this, we first need to standardize the given equation to reveal its structure.
Divide each term by 192 to normalize the equation:
16x²/192 - 12y²/192 = 192/192 x²/12 - y²/16 = 1
This represents a hyperbola that is centered at the origin (0,0) with its transverse axis along the x-axis and opens horizontally.
• The vertices are found by taking the square root of the denominator under the x term, which gives us ±√12. Therefore, the vertices are at (3,0) and (-3,0).
• The foci are calculated using the formula c² = a² + b², where c is the distance from the center to a focus, and a and b are the square roots of the denominators under the x and y terms, respectively. After finding c, the coordinates of the foci will be (±c,0).
• The asymptotes of a hyperbola are given by y = ±(b/a)x for hyperbolas that open horizontally. Substituting the values of a and b, we get the equations of the asymptotes as y = ±(4/3.464)x.
Using these methods, students can identify the important features of a hyperbola, which can be useful as graphing exercises or in applications involving hyperbolic geometry.