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Q)A parabola is drawn to pass through A & B, the ends of a diameter of a given circle of of radius 'a' and to have a directrix a tangent to a concentric circle of radius 'b' the axes being AB and a perpendicular diameter. Prove that the locus of the focus of the parabola is x2b2+y2b2-a2=1
please give the detailed solution.
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10 Answers
1saikat the locus that u have given is not a parabola, i am getting a very different answer, please recheck ur question
11someone elaborate this question.either my english is poor or saikat your language is unclear to me.
1The question is obviously correct, see S.L. LONEY for confirmation and Redion, the locus may be any conic in the world, it doesn't matter.
21i took AB as X axis.... and perpendicular diameter as y axis
let the equation of tangent or directrix be y=mx+c
then we have c2=b2(1+m2)
now the axis of parabola is perpendicular to directrix and passing through origin....(because axis is nothing but normal)
and the focus has to lie on the normal ..let the co ordinate of focus be (h,k)
so we have m=-h/k
as (a,0) and (-a,0) lies on the parabola...we can apply SP=PM...
but i am not getting the answer, where my approach is wrong??
23let A,B ≡(-a,0) and (a,0)
let focus = h,k
let directrix ≡ mx-y+c=0
now c2 = b2(1+m2)
and by definition of parabola
(a-h)2 + k2 = (ma + c )2(1+m2)
and
(a+h)2 + k2 = ( c - am )2(1+m2)
subtract eqn 1 frm 2
(a+h)2 - (a-h)2= ( c - am )2(1+m2) - (ma + c )2(1+m2)
hence h = - b2mc
now
(a+h)2 + k2 = ( c - am )2(1+m2) = c2 ( 1 - amc )2(1+m2) = b2 ( 1 - amc )2
= b2 ( 1 + ahb2 )2
on simplifying we get
h2b2 + k2b2-a2 = 1