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The center \( I \) of the hyperbola \( (H) \) is at \( (1, \frac{m}{3}) \). For all real values of \( m \) except \( m = 3 \), the hyperbola \( (H) \) intersects the line \( (D) \) at 2 points.
Solución
To solve the problem, we will follow the steps outlined in the prompt.
### Step 1: Find the coordinates of the center \( I \) of the hyperbola \( (H) \)
The function given is:
\[
f(x) = \frac{m x}{3 x - 3} - m
\]
To find the center of the hyperbola, we need to rewrite the function in a standard form. First, we simplify \( f(x) \):
1. Combine the terms:
\[
f(x) = \frac{m x}{3(x - 1)} - m = \frac{m x - 3m(x - 1)}{3(x - 1)} = \frac{m x - 3m x + 3m}{3(x - 1)} = \frac{-2m x + 3m}{3(x - 1)}
\]
2. The hyperbola can be expressed in the form \( \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \) where \( (h, k) \) is the center.
To find the center, we need to identify the vertical and horizontal asymptotes of the hyperbola. The vertical asymptote occurs when the denominator is zero:
\[
3x - 3 = 0 \implies x = 1
\]
The horizontal asymptote can be found by analyzing the behavior of \( f(x) \) as \( x \to \infty \):
\[
f(x) \approx \frac{m x}{3 x} = \frac{m}{3}
\]
Thus, the horizontal asymptote is \( y = \frac{m}{3} \).
The center \( I \) of the hyperbola is at the intersection of the asymptotes:
\[
I = (1, \frac{m}{3})
\]
### Deduce the set of points \( I \) when \( m \) varies
As \( m \) varies, the \( y \)-coordinate of the center \( I \) changes. Therefore, the set of points \( I \) is:
\[
I = \{ (1, \frac{m}{3}) \mid m \in \mathbb{R}, m \neq 3 \}
\]
This describes a vertical line at \( x = 1 \) for all \( y \) values except \( y = 1 \) (when \( m = 3 \)).
### Step 2: Study the number of points of intersection of \( (H) \) and \( (D) \)
The line \( (D) \) is given by:
\[
y = -x + 4
\]
To find the points of intersection, we set \( f(x) = -x + 4 \):
\[
\frac{m x}{3 x - 3} - m = -x + 4
\]
Rearranging gives:
\[
\frac{m x}{3 x - 3} + x - 4 - m = 0
\]
Multiplying through by \( 3x - 3 \) to eliminate the fraction:
\[
m x + (x - 4 - m)(3x - 3) = 0
\]
Expanding this:
\[
m x + (3x^2 - 3x - 12x + 12 - 3m) = 0
\]
Combining like terms:
\[
3x^2 + (m - 15)x + (12 - 3m) = 0
\]
This is a quadratic equation in \( x \). The number of solutions (points of intersection) depends on the discriminant \( \Delta \):
\[
\Delta = (m - 15)^2 - 4 \cdot 3 \cdot (12 - 3m)
\]
Calculating the discriminant:
\[
\Delta = (m - 15)^2 - 12(12 - 3m) = (m - 15)^2 - 144 + 36m
\]
Simplifying:
\[
\Delta = m^2 - 30m + 225 - 144 + 36m = m^2 + 6m + 81
\]
Now, we analyze the discriminant \( \Delta \):
1. **If \( \Delta > 0 \)**: There are 2 points of intersection.
2. **If \( \Delta = 0 \)**: There is 1 point of intersection (tangential).
3. **If \( \Delta < 0 \)**: There are no points of intersection.
Since \( m^2 + 6m + 81 \) is a quadratic with a positive leading coefficient and its discriminant is negative (as \( 6^2 - 4 \cdot 1 \cdot 81 < 0 \)), it is always positive for all real \( m \).
Thus, \( \Delta > 0 \) for all \( m \in \mathbb{R} \), meaning there are always 2 points of intersection between the hyperbola \( (H) \) and the line \( (D) \) for all values of \( m \) except \( m = 3 \).
### Conclusion
1. The coordinates of the center \( I \) of the hyperbola \( (H) \) are \( (1, \frac{m}{3}) \), forming a vertical line at \( x = 1 \) as \( m \) varies.
2. The hyperbola \( (H) \) intersects the line \( (D) \) at 2 points for all \( m \in \mathbb{R} \) except \( m = 3 \).
Revisado y aprobado por el equipo de tutoría de UpStudy
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