Tangents And Normals

Step 1: Parametrize the curve.

The equation of the curve is . We can parametrize it using and .

Step 2: Find dy/dx.

First, find and :

Now, find :

Step 3: Find the equation of the tangent line.

The equation of the tangent line at the point is:

Step 4: Find the x and y intercepts.

To find the x-intercept, set :

To find the y-intercept, set :

Step 5: Calculate the length of the intercept.

The length of the tangent intercepted between the axes is the distance between the points and , which is:

Final Answer

Since , the length of the tangent intercepted between the axes is constant.

1. Tangent to the Ellipse:
Given ellipse: A point on the ellipse:
Equation of the tangent at this point:

2. Auxiliary Circle:
The auxiliary circle corresponding to the ellipse is: We are to find the points where the tangent line (Equation I) intersects this circle.

3. Joint Equation of Lines from Origin:
Let the points of intersection of the tangent with the circle be and . The joint equation of lines joining the origin (0,0) to and is:

4. Condition for Lines to be Perpendicular:
The condition for two lines from the origin to be perpendicular is: Let’s expand the right-hand side: So the complete equation is: Now move all terms to one side: For the lines to be perpendicular, the sum of coefficients of and must be zero: This leads to a relationship that can be manipulated to find **eccentricity**, or further simplified in specific cases (like special angles) to derive other properties.

5. Final Result:
The value involving area or geometry from this analysis leads to the final answer (in this case) being:
Option (B):