CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A focal chord of the parabola y2=4ax meets it at P and Q. If S is the focus then 1SP+1SQ= 
  1. a
  2. 2a


Solution

The correct option is A
Let P=(at21,2at1),Q=(at22,2at2) Now focus S=(a,0)
Since PQ is a focal chord,t1t2=1
SP=(at21a)2+(2at10)2=a(t211)2+4t21=a(t211)2=a(t211)
SQ=(at22a)2+(2at20)2=a(t221)2+4t21=a(t221)2=a(t221)
1SP+1SQ=1a(t21+1)+1a(t22+1)=1a[1t21+1+1t22+1]=1a[t22+1+t22+1(t22+1)(t22+1)+]=1a(t21+t22+2t21+t21t22+t22+1)
=1a[t21+t22+21+t21+t22+1]=1a

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image