Let's solve the problem step by step.
We need to find the product of the distances from the foci of the hyperbola [tex]\(\frac{x^2}{128} - \frac{y^2}{18} = 1\)[/tex] to any tangent of this hyperbola.
1. Identify the form of the hyperbola:
The given equation of the hyperbola is [tex]\(\frac{x^2}{128} - \frac{y^2}{18} = 1\)[/tex].
This is in the standard form [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)[/tex], where [tex]\(a^2 = 128\)[/tex] and [tex]\(b^2 = 18\)[/tex].
2. Calculate [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[
a = \sqrt{128} = 11.3137 \quad (\text{approx})
\][/tex]
[tex]\[
b = \sqrt{18} = 4.2426 \quad (\text{approx})
\][/tex]
3. Distance between the foci:
For a hyperbola, the distance [tex]\(2c\)[/tex] between the foci is determined by the relationship [tex]\(c^2 = a^2 + b^2\)[/tex].
[tex]\[
c = \sqrt{a^2 + b^2} = \sqrt{128 + 18} = \sqrt{146} = 12.083 \quad (\text{approx})
\][/tex]
4. Product of the distances from the foci to any tangent:
For a hyperbola [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)[/tex], the product of the distances from the foci to any tangent is [tex]\(b^2\)[/tex].
Therefore, the product of the distances from the foci to any tangent to the hyperbola [tex]\(\frac{x^2}{128} - \frac{y^2}{18} = 1\)[/tex] is:
[tex]\[
b^2 = 18
\][/tex]
Thus, the product of the distances from the foci of [tex]\(\frac{x^2}{128} - \frac{y^2}{18} = 1\)[/tex] to any tangent to it is [tex]\(18\)[/tex].
