Let `C : x^(2) + y^(2) = 9, E : (x^(2))/(9) + (y^(2))/(4) =1` and L : y = 2x be three curves P be a point on C and PL be the perpendicular to the major axis of ellipse E. PL cuts the ellipse at point M.
If equation of normal to C at point P be `L : y = 2x` then the equation of the tangent at M to the ellipse E is

Options

A. `x + 3y pm 3sqrt(5)= 0`
B. `4x + 3y pm sqrt(5) = 0`
C. `x + y pm 3 = 0`
D. None of these

Answer

To solve the problem step by step, we will break down the information given and apply the necessary mathematical concepts.

### Step 1: Identify the curves and their equations
We have the following curves:
1. Circle \( C: x^2 + y^2 = 9 \)
2. Ellipse \( E: \frac{x^2}{9} + \frac{y^2}{4} = 1 \)
3. Line \( L: y = 2x \)

### Step 2: Find the point \( P \) on the circle \( C \)
Let \( P(a, b) \) be a point on the circle \( C \). Since \( P \) lies on the circle, it satisfies the equation:
\[ a^2 + b^2 = 9 \]

### Step 3: Determine the slope of the normal at point \( P \)
The normal to the circle at point \( P(a, b) \) has a slope that is the negative reciprocal of the slope of the tangent. The slope of the radius to point \( P \) is given by:
\[ \text{slope of radius} = \frac{b}{a} \]
Thus, the slope of the normal line \( L \) is:
\[ \text{slope of normal} = -\frac{a}{b} \]

### Step 4: Set the normal line equal to the given line \( L \)
Since the equation of the normal line is given as \( y = 2x \), we can equate the slopes:
\[ -\frac{a}{b} = 2 \implies a = -2b \]

### Step 5: Substitute \( a \) in the circle equation
Now substitute \( a = -2b \) into the circle equation:
\[ (-2b)^2 + b^2 = 9 \implies 4b^2 + b^2 = 9 \implies 5b^2 = 9 \implies b^2 = \frac{9}{5} \implies b = \pm \frac{3}{\sqrt{5}} \]
Now substituting back to find \( a \):
\[ a = -2b = -2\left(\pm \frac{3}{\sqrt{5}}\right) = \mp \frac{6}{\sqrt{5}} \]
Thus, the coordinates of point \( P \) can be:
\[ P\left(-\frac{6}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right) \quad \text{or} \quad P\left(\frac{6}{\sqrt{5}}, -\frac{3}{\sqrt{5}}\right) \]

### Step 6: Find the perpendicular line \( PL \)
The line \( PL \) is perpendicular to the major axis of the ellipse \( E \) (the x-axis). Thus, it has the form \( x = k \) where \( k \) is the x-coordinate of point \( P \).

### Step 7: Find the point \( M \) where \( PL \) intersects the ellipse
Substituting \( x = -\frac{6}{\sqrt{5}} \) into the ellipse equation:
\[ \frac{\left(-\frac{6}{\sqrt{5}}\right)^2}{9} + \frac{y^2}{4} = 1 \]
Calculating:
\[ \frac{\frac{36}{5}}{9} + \frac{y^2}{4} = 1 \implies \frac{4}{5} + \frac{y^2}{4} = 1 \implies \frac{y^2}{4} = 1 - \frac{4}{5} = \frac{1}{5} \implies y^2 = \frac{4}{5} \implies y = \pm \frac{2}{\sqrt{5}} \]
Thus, the point \( M \) is:
\[ M\left(-\frac{6}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right) \quad \text{or} \quad M\left(-\frac{6}{\sqrt{5}}, -\frac{2}{\sqrt{5}}\right) \]

### Step 8: Find the equation of the tangent at point \( M \)
The slope of the tangent line to the ellipse at point \( M \) can be found using the derivative of the ellipse equation. The implicit differentiation gives:
\[ \frac{d}{dx}\left(\frac{x^2}{9} + \frac{y^2}{4}\right) = 0 \implies \frac{2x}{9} + \frac{2y}{4}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{9y} \]
Substituting \( M\left(-\frac{6}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right) \):
\[ \frac{dy}{dx} = -\frac{4\left(-\frac{6}{\sqrt{5}}\right)}{9\left(\frac{2}{\sqrt{5}}\right)} = \frac{24/\sqrt{5}}{18/\sqrt{5}} = \frac{24}{18} = \frac{4}{3} \]

### Step 9: Write the equation of the tangent line
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \implies y - \frac{2}{\sqrt{5}} = \frac{4}{3}\left(x + \frac{6}{\sqrt{5}}\right) \]
Simplifying:
\[ y = \frac{4}{3}x + \frac{8}{\sqrt{5}} + \frac{2}{\sqrt{5}} = \frac{4}{3}x + \frac{10}{\sqrt{5}} = \frac{4}{3}x + 2\sqrt{5} \]

### Final Answer
The equation of the tangent at point \( M \) to the ellipse \( E \) is:
\[ y = \frac{4}{3}x + 2\sqrt{5} \]