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JEE Main 2023
Conic Sections
Parabola
Hard

Question

The focus of the parabola y2=4x+16y^2=4 x+16 is the centre of the circle CC of radius 5 . If the values of λ\lambda, for which C passes through the point of intersection of the lines 3xy=03 x-y=0 and x+λy=4x+\lambda y=4, are λ1\lambda_1 and λ2,λ1<λ2\lambda_2, \lambda_1<\lambda_2, then 12λ1+29λ212 \lambda_1+29 \lambda_2 is equal to ________ .

Answer: 2

Solution

This problem is a comprehensive test of analytical geometry, requiring a strong grasp of parabolas, circles, and straight lines. The solution involves a sequence of steps: first, identifying the key properties of the parabola to find the center of the circle; second, defining the circle's equation; third, finding the intersection point of two lines in terms of a parameter λ\lambda; and finally, using the condition that this point lies on the circle to solve for λ\lambda.


1. Key Concepts and Formulas

  • Parabola: The standard form of a parabola opening to the right with its vertex at the origin is y2=4axy^2 = 4ax. Its focus is at (a,0)(a,0). If the parabola is shifted, its equation becomes $(y-k)^2 =

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