SOLUTION: The equation of hyper bola is ((X)^2/9)-((Y)^2/16)=1 And it's asymptotes are y=4x/3 and y=-4x/3 Here, h(x) represents the portion of the hyperbola in the first quadrant. Based on

Question 1199700: The equation of hyper bola is ((X)^2/9)-((Y)^2/16)=1
And it's asymptotes are y=4x/3 and y=-4x/3
Here, h(x) represents the portion of the hyperbola in the first quadrant. Based on this:
a. Write an expression for h(x) in terms of x .
b. Use the expression from part a. to justify why h(x) lies below the line y=4x/3 .
Answer by textot(100) (Show Source): You can put this solution on YOUR website!
**a. Find the expression for h(x)**
1. **Isolate y^2:**
- (x^2)/9 - (y^2)/16 = 1
- (y^2)/16 = (x^2)/9 - 1
- y^2 = 16 * [(x^2)/9 - 1]
- y^2 = (16x^2)/9 - 16
2. **Solve for y:**
- y = ±√[(16x^2)/9 - 16]
Since we're considering only the portion of the hyperbola in the first quadrant (where both x and y are positive), we take the positive square root:
- h(x) = √[(16x^2)/9 - 16]
**b. Justify why h(x) lies below the line y = 4x/3**
1. **Consider the equation of the asymptote:**
- The asymptote in the first quadrant is y = 4x/3.
2. **Analyze the behavior of h(x):**
- h(x) represents the y-coordinate of the hyperbola for a given x-value.
- As x increases, the value of (16x^2)/9 increases significantly.
- However, the square root function grows relatively slowly compared to a linear function like y = 4x/3.
3. **Conclusion:**
- The square root in h(x) will cause the y-values of the hyperbola to increase at a slower rate than the linear increase of the asymptote y = 4x/3.
- Therefore, the graph of h(x) will always remain below the line y = 4x/3 in the first quadrant.
**In essence:**
* The hyperbola approaches the asymptote y = 4x/3 as x increases, but it never touches or crosses it.
* The square root in the expression for h(x) limits the growth of the y-values compared to the linear growth of the asymptote.
This analysis demonstrates why the portion of the hyperbola in the first quadrant (h(x)) lies below the asymptote y = 4x/3.

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