Q is a point on the auxiliary circle of an ellipse. P is the corresponding point on ellipse. N is the foot of perpendicular from focus S, to the tangent of auxiliary circle at Q. Then

Options

A. SP=SN
B. SP=PQ
C. PN=SP
D. NQ=SP

Answer

To solve the problem, we need to establish the relationships between the points Q, P, S, and N as described in the question. Let's break down the solution step by step.

### Step 1: Understand the Ellipse and Auxiliary Circle
The standard equation of an ellipse is given by:
\[ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \]
where \( A > B \). The auxiliary circle of the ellipse has a radius equal to \( A \) and is centered at the origin.

### Step 2: Identify Points
1. **Focus (S)**: The foci of the ellipse are located at \( S = (\pm C, 0) \), where \( C = \sqrt{A^2 - B^2} \).
2. **Point on Ellipse (P)**: A point \( P \) on the ellipse can be represented as:
\[ P = (A \cos \theta, B \sin \theta) \]
3. **Point on Auxiliary Circle (Q)**: The corresponding point \( Q \) on the auxiliary circle is given by:
\[ Q = (A \cos \theta, A \sin \theta) \]

### Step 3: Find the Tangent at Point Q
The tangent to the auxiliary circle at point \( Q \) can be determined using the point-slope form of the tangent line. The equation of the tangent line at point \( Q \) is:
\[ x \cos \theta + y \sin \theta = A \]

### Step 4: Find the Foot of the Perpendicular (N)
To find the foot of the perpendicular from the focus \( S \) to the tangent line, we can use the formula for the distance from a point to a line. The distance \( d \) from point \( S = (Ae, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Substituting the values, we can find the coordinates of point \( N \).

### Step 5: Calculate Distances SP and SN
1. **Distance \( SP \)**:
\[ SP = \sqrt{(Ae - A \cos \theta)^2 + (0 - B \sin \theta)^2} \]
Simplifying this gives:
\[ SP = \sqrt{A^2 e^2 - 2A^2 e \cos \theta + A^2 \cos^2 \theta + B^2 \sin^2 \theta} \]
Using \( B^2 = A^2 - C^2 \), we can simplify further.

2. **Distance \( SN \)**:
The distance \( SN \) is calculated similarly using the coordinates of \( S \) and \( N \).

### Step 6: Establish Relationships
After calculating both distances, we find that:
\[ SP = SN \]
This shows that the distance from the focus \( S \) to the point \( P \) on the ellipse is equal to the distance from \( S \) to the foot of the perpendicular \( N \) on the tangent line at point \( Q \).

### Conclusion
Thus, we conclude that:
\[ SP = SN \]