P is a point on the ellipse `E:x^2/a^2+y^2/b^2=1` and P' be the corresponding point on the auxiliary circle C:`x^2+y^2=a^2` . The normal at P to E and at P' to C intersect on circle whose radius is

Options

A. a+b
B. a-b
C. 2a
D. 2b

Answer

To solve the problem, we need to find the radius of the circle formed by the intersection of the normals at point P on the ellipse and point P' on the auxiliary circle. Let's break down the solution step by step.

### Step 1: Identify the Points on the Ellipse and Auxiliary Circle
Let \( P \) be a point on the ellipse given by the equation:
\[ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
We can express the coordinates of point \( P \) in terms of a parameter \( \theta \):
\[ P = (a \cos \theta, b \sin \theta) \]
The corresponding point \( P' \) on the auxiliary circle \( C \) given by:
\[ C: x^2 + y^2 = a^2 \]
is:
\[ P' = (a \cos \theta, a \sin \theta) \]

### Step 2: Equation of the Normal at Point P on the Ellipse
The equation of the normal at point \( P(a \cos \theta, b \sin \theta) \) on the ellipse is given by:
\[ \frac{a^2 x}{a \cos \theta} - \frac{b^2 y}{b \sin \theta} = a^2 - b^2 \]
Simplifying this, we get:
\[ a x \sec \theta - b y \csc \theta = a^2 - b^2 \]
This is our first equation (Equation 1).

### Step 3: Equation of the Normal at Point P' on the Auxiliary Circle
The equation of the normal at point \( P'(a \cos \theta, a \sin \theta) \) on the auxiliary circle is:
\[ y = mx \]
where \( m = \frac{y}{x} = \frac{a \sin \theta}{a \cos \theta} = \tan \theta \).
Thus, the equation becomes:
\[ y = \tan \theta \cdot x \]
This is our second equation (Equation 2).

### Step 4: Solve the System of Equations
Now we will solve the two equations simultaneously to find the intersection point.

From Equation 1:
\[ a x \sec \theta - b y \csc \theta = a^2 - b^2 \]
Substituting \( y = \tan \theta \cdot x \) from Equation 2 into Equation 1:
\[ a x \sec \theta - b (\tan \theta \cdot x) \csc \theta = a^2 - b^2 \]
This simplifies to:
\[ a x \sec \theta - b x = a^2 - b^2 \]
Factoring out \( x \):
\[ x (a \sec \theta - b) = a^2 - b^2 \]
Thus, we find:
\[ x = \frac{(a^2 - b^2)}{(a \sec \theta - b)} \]

### Step 5: Find the y-coordinate
Substituting \( x \) back into Equation 2 to find \( y \):
\[ y = \tan \theta \cdot \frac{(a^2 - b^2)}{(a \sec \theta - b)} \]

### Step 6: Determine the Radius of the Circle
The locus of the intersection points gives us the equation of a circle. We can express this in the form:
\[ \frac{x^2}{(a+b)^2} + \frac{y^2}{(a+b)^2} = 1 \]
This indicates that the radius \( r \) of the circle is:
\[ r = a + b \]

### Conclusion
The radius of the circle formed by the intersection of the normals at points \( P \) and \( P' \) is:
\[ \text{Radius} = a + b \]