IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

varun2275

PQ is a variable chord of x^2/a^2 +y^2/b^2 = 1. if PQ subtends a rt angle at the centre of the ellipse, then find 1/OP^2 + 1/OQ^2 ( O is the origin)

PLEEEEEEEASE some one help asap!!!

varun2275

options are :
a) 1/a^2 + 1/b^2 b) 2/a^2 + 1/b^2 c) 1/a^2 + 2/b^2 c) 2( 1/a^2 + 1/b^2)

ankurgupta91

the variable chord PQ subtends riight angle at centre i.e origin O
it must be along X-axis or Y-axis.

if it is along X-axis then OP=a and OQ=a
thus , 1/OP^2 + 1/OQ^2 = 2/a^2

and if it along Y-axis then OP=b and OQ=b
thus, 1/OP^2 + 1/OQ^2 =2/b^2

thats the answer
thanks............

varun2275

huh? i didnt get u, how r OP and OQ = a in the first case and = b in the second?? can u explain?? and its not mentioned that the chord is along the x axis or y axis......

ankurgupta91

sry i understoof the ques in other way
nw let P(acos#,bsin#)
then Q be(acos(90+#),bsin(90+#))
i.e Q(-asin#,bcos#)
nw slope of OP *slope of OQ =-1
i.e (bsin#/acos#)*(-bcos#/asin#)=-1
i.e b^2=a^2
Nw distance OP^2= a^2cos#^2 + b^2sin#^2
=a^2(cos#^2+sin#^2)
=a^2
simlarly OQ^2=a^2sin#^2+b^2cos#^2
=b^2
1/OP^2 +1/OQ^2 = 1/a^2+1/b^2
the ans. is (a)
thanx...............

akella.as

did not understand q properly

cbg2007

varun i will tell the method but i do not give the solution.but don't forget me to rate.
here is the solution.

you just homisinise the chord equation with the chord equation.
then coefficient of x^2+coefficient of y^2 =0

sushu

yes it must be homoginised.

but ankur is also rite.iam not sure.

sv_sadhu

ANSWER IS 1/a^2+1/b^2

Ask & Discuss Questions with Community & Experts