If a ray of light incident along the line `3x+(5-4sqrt(2))y=15` gets reflected from the hyperbola `(x^2)/(16)-(y^2)/9=1` , then its reflected ray goes along the line. `xsqrt(2)-y+5=0` (b) `sqrt(2)y-x+5=0` `sqrt(2)y-x-5=0` (d) none of these

Options

A. `xsqrt2-y+5=0`
B. `sqrt2y-x+5=0`
C. `sqrt2y-x-5=0`
D. none of these

Explanation

To solve the problem step by step, we will analyze the given information and derive the equation of the reflected ray from the hyperbola.

### Step 1: Identify the hyperbola and its properties
The hyperbola is given by the equation:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]
From this, we can identify the semi-major axis \(a = 4\) and semi-minor axis \(b = 3\). The eccentricity \(e\) can be calculated as:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]
The foci of the hyperbola are located at \((\pm ae, 0) = (\pm 5, 0)\).

### Step 2: Identify the incident ray
The incident ray is given by the equation:
\[ 3x + (5 - 4\sqrt{2})y = 15 \]
We can rewrite this in slope-intercept form \(y = mx + c\) to find the slope:
\[ y = \frac{-3}{4\sqrt{2} - 5}x + \frac{15}{5 - 4\sqrt{2}} \]
The slope \(m_i\) of the incident ray is:
\[ m_i = \frac{-3}{5 - 4\sqrt{2}} \]

### Step 3: Determine the point of intersection with the hyperbola
We will find the point of intersection \(P\) of the incident ray with the hyperbola. We can parameterize a point on the hyperbola using:
\[ P = (4\sec\theta, 3\tan\theta) \]
Substituting \(x = 4\sec\theta\) and \(y = 3\tan\theta\) into the line equation gives us:
\[ 3(4\sec\theta) + (5 - 4\sqrt{2})(3\tan\theta) = 15 \]
This simplifies to:
\[ 12\sec\theta + (5 - 4\sqrt{2})3\tan\theta = 15 \]

### Step 4: Solve for \(\theta\)
Rearranging the equation:
\[ (5 - 4\sqrt{2})3\tan\theta = 15 - 12\sec\theta \]
This can be solved for \(\theta\). Notably, we find that \(\theta = \frac{\pi}{4}\) satisfies the equation.

### Step 5: Find the coordinates of point \(P\)
Substituting \(\theta = \frac{\pi}{4}\):
\[ P = (4\sec(\frac{\pi}{4}), 3\tan(\frac{\pi}{4})) = (4\sqrt{2}, 3) \]

### Step 6: Determine the reflected ray
The reflected ray will pass through the point \(P\) and the second focus \(S'(-5, 0)\). The slope of the line joining \(P\) and \(S'\) is:
\[ m_r = \frac{3 - 0}{4\sqrt{2} + 5} \]
Using point-slope form, the equation of the reflected ray is:
\[ y - 0 = \frac{3}{4\sqrt{2} + 5}(x + 5) \]

### Step 7: Simplify the equation of the reflected ray
Rearranging gives:
\[ y = \frac{3}{4\sqrt{2} + 5}(x + 5) \]
This can be further simplified to find the final equation of the reflected ray.

### Step 8: Check against options
After simplifying, we compare the derived equation with the given options to determine the correct answer.