If the normal at any point P on ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =1` meets the auxiliary circle at Q and R such that `/_QOR = 90^(@)` where O is centre of ellipse, then

Options

A. (a) `a^(4) +2b^(3) ge 3a^(2)b^(2)`
B. (b) `a^(4) +2b^(4) ge 5a^(2)b^(2)+2a^(3)b`
C. (c) `a^(4)+2b^(4) ge 3a^(2)b^(2)+ab`
D. (d) None of these

Explanation

To solve the problem, we need to analyze the given ellipse and the conditions provided. Here’s a step-by-step solution:

### Step 1: Understand the Ellipse and its Normal
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The parametric coordinates of a point \( P \) on the ellipse can be expressed as:
\[ P(a \cos \theta, b \sin \theta) \]

### Step 2: Write the Equation of the Normal
The equation of the normal at point \( P \) can be derived using the slope of the tangent. The slope of the tangent at point \( P \) is given by:
\[ \frac{dy}{dx} = -\frac{b^2 \cos \theta}{a^2 \sin \theta} \]
Thus, the slope of the normal is the negative reciprocal:
\[ \text{slope of normal} = \frac{a^2 \sin \theta}{b^2 \cos \theta} \]
Using the point-slope form of the equation of a line, the equation of the normal at point \( P \) is:
\[ y - b \sin \theta = \frac{a^2 \sin \theta}{b^2 \cos \theta} (x - a \cos \theta) \]

### Step 3: Auxiliary Circle
The auxiliary circle of the ellipse has the equation:
\[ x^2 + y^2 = a^2 \]

### Step 4: Intersection of Normal and Auxiliary Circle
To find the points \( Q \) and \( R \) where the normal intersects the auxiliary circle, we substitute the equation of the normal into the equation of the circle.

### Step 5: Condition for Angle \( \angle QOR = 90^\circ \)
For the angle \( \angle QOR \) to be \( 90^\circ \), the condition is that the coefficients of \( x^2 \) and \( y^2 \) in the resulting equation must satisfy:
\[ \text{Coefficient of } x^2 + \text{Coefficient of } y^2 = 1 \]

### Step 6: Simplification
After substituting and simplifying, we find that:
\[ 1 - \frac{a^4 \sec^2 \theta}{a^2 - b^2} + 1 - \frac{b^4 \csc^2 \theta}{a^2 - b^2} = 0 \]
This leads to the condition:
\[ 2 - \left( \frac{a^4 \sec^2 \theta + b^4 \csc^2 \theta}{a^2 - b^2} \right) = 0 \]

### Step 7: Rearranging and Applying AM-GM Inequality
We rearrange the terms and apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ a^4 + 2b^4 - 5a^2b^2 \geq 2a^3b \]
This gives us the necessary condition for the angles formed by the intersection points.

### Conclusion
The final condition derived from the above steps leads us to the conclusion that the correct option is:
\[ \text{Option B: } a^4 + 2b^4 - 5a^2b^2 \geq 2a^3b \]