ParabolamediumFree

Find a if Normals Meet on y²=4ax for Chord lx+my=1 | JEE

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

Question
Let the line lx+my=1lx + my = 1 intersect the parabola y2=4axy^2 = 4ax at points PP and QQ. The normals at PP and QQ meet at the point RR. If RR lies on the parabola y2=4axy^2 = 4ax, then aa equals:
A1l\dfrac{1}{l}
B12l\dfrac{1}{2l}
C−1l-\dfrac{1}{l}
D−12l-\dfrac{1}{2l}correct
Solution
Step 1: A point (at2,2at)(at^2, 2at) on lx+my=1lx + my = 1 gives alt2+2amt−1=0alt^2 + 2amt - 1 = 0, so the parameters of P,QP, Q satisfy t1t2=−1alt_1 t_2 = -\dfrac{1}{al}. Step 2: Normals at t1,t2t_1, t_2 meet on the parabola only if t1t2=2t_1 t_2 = 2. Therefore
−1al=2⇒a=−12l.-\frac{1}{al} = 2 \Rightarrow a = -\frac{1}{2l}.
 Correct answer: (4)
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