Q1: Consider the lines L1 and L2 defined by
For a fixed constant λ, let C be the locus of a point P such that the product of the distance of P from L1 and the distance of P from L2 is λ2. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is √270. Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'. ]
The value of λ2 is __________. [JEE Advanced 2021 Paper 1]
Ans: 9
According to the question,
Let R ≡ (x1, y1) and S(x2, y2)
∵ C cuts y − 1 = 2x at R and S.
So,
∵ RS2 = 270 (given)
Q2: Consider the lines L1 and L2 defined by
For a fixed constant λ, let C be the locus of a point P such that the product of the distance of P from L1 and the distance of P from L2 is λ2. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is √270. Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'. ]
The value of D2 is __________. [JEE Advanced 2021 Paper 1]
Ans: 77.14
According to the question,
Let R ≡ (x1, y1) and S(x2, y2)
∵ C cuts y − 1 = 2x at R and S.
So,
∵ RS2 = 270 (given)
Now, mid-point of RS is and slope of RS = 2 and slope of
On solving x + 2y − 2 = 0 with C, we get
Hence,
=
209 videos|443 docs|143 tests
|
1. What are some important concepts to understand in straight lines for JEE Advanced? |
2. How can I find the equation of a line knowing its slope and a point it passes through? |
3. What is the condition for two lines to be parallel? |
4. How can I find the distance between a point and a line? |
5. Can you explain the intercept form of a straight line equation? |
209 videos|443 docs|143 tests
|
|
Explore Courses for JEE exam
|