ParabolaeasyFree

Range of t₁t₂ for Chord x-b+λy=0 of y²=4ax | JEE

JEE Maths question with a full step-by-step solution.

Question
The line x−b+λy=0x - b + \lambda y = 0 cuts the parabola y2=4axy^2 = 4ax (a>0)(a > 0) at P(t1)P(t_1) and Q(t2)Q(t_2). If b∈[2a,4a]b \in [2a, 4a], then the range of t1t2t_1 t_2, where λ∈R\lambda \in \mathbb{R}, is:
A[−4,−2][-4, -2]correct
B[2,4][2, 4]
C[4,16][4, 16]
D[−16,−4][-16, -4]
Solution
Step 1: The chord P(t1)Q(t2)P(t_1)Q(t_2) is x−t1+t22y+at1t2=0x - \dfrac{t_1 + t_2}{2}y + at_1 t_2 = 0. Matching x+λy−b=0x + \lambda y - b = 0 gives b=−at1t2b = -at_1 t_2. Step 2: With b∈[2a,4a]b \in [2a, 4a] and a>0a > 0,
2≤−t1t2≤4⇒t1t2∈[−4,−2].2 \le -t_1 t_2 \le 4 \Rightarrow t_1 t_2 \in [-4, -2].
 Correct answer: (1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Parabola · easy
The Cartesian equation of the curve whose parametric equations are x=t2+2t+3x = t^2 + 2t + 3 and y=t+1y = t + 1 is:
Parabola · easy
If x+a=0x + a = 0 is the directrix of the parabola y2=2y+ax+2y^2 = 2y + ax + 2, then the value of aa is:
Parabola · medium
The length of the latus rectum of the parabola x=t2+t+1x = t^2 + t + 1, y=t2+2t+3y = t^2 + 2t + 3 is:
Parabola · medium
If any point P(x,y)P(x, y) which satisfies the relation (5x−1)2+(5y−2)2=λ(3x−4y−1)2(5x - 1)^2 + (5y - 2)^2 = \lambda(3x - 4y - 1)^2 represents a parabola, then:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath