IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - Parabola-4

kane

PQ is any chord of y²=4ax whose right bisector meets the axis in M and the ordinate of mid pt. of PQ meets the axis in L,then prove LM is constant = semi latus rectum of the parabola

sorry folks no figure given plz make one here if you can i will definitely rate that one too

budku007

not able to understand this line

and the ordinate of mid pt. of PQ meets the axis

kane

me too

cursed_mask

Assume co-ordinates of mid-point of PQ as (h,k) ...u can then write the equation of the chord bisected at the point (h,k) (the equation of PQ) from T=S. NOw take perpendicular bisector of this chord through (h,k) and find its intersection with the x-axis...the ordinate of the mid-point is already known as k....now u can find the required length..and show it is independent of PQ or (h,k)

rhd92781

take any two points on the parabola (at1^2, 2at1) n (at2^2, 2at2)

mid point = (a(t1^2+t2^2)/2, a(t1+t2))

slope of chord joining dis points = 2/(t1+t2)

so, u can find the eqn of right bisector as it passes through its mid point and having slope = -(t1+t2)/2

it comes

2y+x(t1+t2)-a(t1+t2)(2+(t1^2+t2^2)/2)=0

it'll cut the major axis (y=0) at (2a+a(t1^2+t2^2)/2, 0) is L

ordinate of mid point will cut the axis at its absiccae

so M=(a(t1^2+t2^2)/2, 0)

now u get LM=2a = half of the latus rectum

now fig needed yaar!

if u still have any doubt or u need fig, nudge me but let me go offline 4 smtime

kane

thnxs to both of you

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