Applications Of Conics

Step 1: Understanding the problem.
The point is given, and circles with variable diameter are touching the given circle internally. The goal is to find the locus of the point , which is the curve . The equation of the given circle is: This is a circle with center and radius 6.
Step 2: Finding the equation of the locus of point B.
The diameter of each circle is , and the point lies on the x-axis. The circle with diameter touches the given circle internally, so the distance from the center of the given circle to the center of the circle with diameter is equal to the difference in their radii. Let the coordinates of point be . The center of the circle with diameter is the midpoint of and . The midpoint of and is: The radius of the circle with diameter is half the distance between and , which is: For the circle to touch the given circle internally, the distance between the centers of the two circles must be equal to the difference in their radii. The center of the given circle is , and the distance from the center of the given circle to the center of the circle with diameter is: This distance is equal to because the given circle has radius 6, and the radius of the circle with diameter is . Thus, we have the equation:
Step 3: Finding the eccentricity of the curve.
The curve is the locus of point , and the equation of this curve is an ellipse. The eccentricity of an ellipse is related to the semi-major axis and the semi-minor axis by the formula: After solving the system of equations, we find that the eccentricity of the curve is .
Step 4: Calculating .
Now that we know , we can calculate : Thus, the final answer is:

Concept: For an ellipse :

• Eccentricity
• Directrices

For hyperbola :

• Length of latus rectum

Step 1: {Find the semi-major axis of the ellipse.} Given directrix: Step 2: {Find the minor axis of the ellipse.} Thus minor axis length: Step 3: {Use hyperbola conditions.} Hyperbola eccentricity . Let hyperbola parameters be . Step 4: {Use latus rectum condition.} Step 5: {Find distance between foci.} For hyperbola: Distance between foci: After simplification with given options:

Step 1: Understanding the Concept:
We parameterize the vertices of the right-angled triangle on the parabola. Using the slope condition for perpendicularity at vertex B, we can relate the parameters of vertices A and C. Then, substitute these relations into the centroid coordinates to find its locus .

Step 2: Key Formula or Approach:
Parametric coordinates for are . Here , so points are .
Slope of a chord joining is .
For perpendicular lines, .
Centroid has coordinates .

Step 3: Detailed Explanation:
Given B , it corresponds to the parameter .
Let A and C have parameters and .
Since angle at B is , the product of slopes of AB and BC is .



Expanding: .
Let the centroid be .
.
.
We know the algebraic identity: .
Substitute the centroid relations and the perpendicularity constraint into this identity:





Multiply by :
.
Replacing with , the locus is the parabola:
.
The length of the latus rectum of this new parabola is the coefficient of the linear term, which is .
We are asked for three times the length of the latus rectum:
.

Step 4: Final Answer:
The required value is 16.