From the focus `(-5,0)` of the ellipse `(x^(2))/(45)+(y^(2))/(20) =1`, a ray of light is sent which makes angle `cos^(-1)((-1)/(sqrt(5)))` with the positive direction of X-axis upon reacting the ellipse the ray is reflected from it. Slope of the reflected ray is

Options

A. `-(3)/(2)`
B. `-(7)/(3)`
C. `-(5)/(4)`
D. `-(2)/(11)`

Explanation

To solve the problem step by step, we will follow the outlined process:

### Step 1: Understand the given ellipse and its properties
The equation of the ellipse is given by:
\[ \frac{x^2}{45} + \frac{y^2}{20} = 1 \]
From this equation, we can identify the semi-major axis \(a\) and semi-minor axis \(b\):
- \(a^2 = 45 \Rightarrow a = \sqrt{45} = 3\sqrt{5}\)
- \(b^2 = 20 \Rightarrow b = \sqrt{20} = 2\sqrt{5}\)

The foci of the ellipse can be calculated using the formula \(c = \sqrt{a^2 - b^2}\):
\[ c = \sqrt{45 - 20} = \sqrt{25} = 5 \]
The foci are located at \((-c, 0)\) and \((c, 0)\), which gives us the coordinates of the foci as \((-5, 0)\) and \((5, 0)\).

### Step 2: Determine the angle of incidence
The angle of the ray with the positive x-axis is given as:
\[ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{5}}\right) \]
From this, we can find the slope of the ray. We know that:
\[ \cos \theta = -\frac{1}{\sqrt{5}} \Rightarrow \text{Using the identity } \sin^2 \theta + \cos^2 \theta = 1 \]
We can find \(\sin \theta\):
\[ \sin^2 \theta = 1 - \left(-\frac{1}{\sqrt{5}}\right)^2 = 1 - \frac{1}{5} = \frac{4}{5} \Rightarrow \sin \theta = \frac{2}{\sqrt{5}} \]
Thus, the slope \(m\) of the incident ray is:
\[ m = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2}{\sqrt{5}}}{-\frac{1}{\sqrt{5}}} = -2 \]

### Step 3: Write the equation of the incident ray
The incident ray passes through the focus \((-5, 0)\) with slope \(-2\). The equation of the line can be written as:
\[ y - 0 = -2(x + 5) \Rightarrow y = -2x - 10 \]

### Step 4: Find the intersection point with the ellipse
To find the intersection point \(P\) of the ray and the ellipse, substitute \(y = -2x - 10\) into the ellipse equation:
\[ \frac{x^2}{45} + \frac{(-2x - 10)^2}{20} = 1 \]
Expanding the second term:
\[ \frac{x^2}{45} + \frac{(4x^2 + 40x + 100)}{20} = 1 \]
Multiplying through by \(180\) (LCM of 45 and 20):
\[ 4x^2 + 36x^2 + 720x + 1800 = 180 \]
Combine like terms:
\[ 40x^2 + 720x + 1620 = 0 \]
Dividing through by 20:
\[ 2x^2 + 36x + 81 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{-36 \pm \sqrt{36^2 - 4 \cdot 2 \cdot 81}}{2 \cdot 2} = \frac{-36 \pm \sqrt{1296 - 648}}{4} = \frac{-36 \pm \sqrt{648}}{4} \]
Calculating \(\sqrt{648} = 18\sqrt{2}\):
\[ x = \frac{-36 \pm 18\sqrt{2}}{4} = -9 \pm \frac{9\sqrt{2}}{2} \]
Choosing the appropriate root that lies within the ellipse, we find:
\[ x = -6 \quad (approx) \]
Substituting back to find \(y\):
\[ y = -2(-6) - 10 = 2 \]
Thus, the coordinates of point \(P\) are \((-6, 2)\).

### Step 5: Determine the slope of the reflected ray
Using the property of reflection in an ellipse, the reflected ray will pass through the other focus \((5, 0)\) and point \(P(-6, 2)\). The slope of the reflected ray \(m_1\) is given by:
\[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 2}{5 - (-6)} = \frac{-2}{11} \]

### Final Answer
The slope of the reflected ray is:
\[ \boxed{-\frac{2}{11}} \]