if d area of d triangle inscribed in d parabola y=4ax wid one vertex @ d origin of d parabola and othr two vertices at d xtremities ofd focal chord is 5a^2/2 then d elngt of d focal chord is
i tried by takin dat point as at^2,2at and then othr end is a/t^2,-2a/t and then 1/2*base*height=area
so i get sum crap in termz of t .........itz tedious .....
othr method plez
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1 Answers
1take the two points t1 and t2.
now, write the area in determinant form....from thr you get a relation between t1 and t2 as ...
t1.t2 (t1-t2)= 5/2 ----------(1)
but we also have ... t1.t2 = -1 ------------(2) {since its a focal chord}
from eqns. 1 n 2,,, t2-t1 =5/2 ---------- (3)
from eqns. 2 n 3,,,, t1+t2= √(25/4 - 4) =3/2 {(a+b)2-(a-b)2=4ab}
now, lenght of the focal chord is
l= √[(at12-at22)2 + (2at1-2at2)2]
= √[a2(t1+t2)2(t1-t2)2 + 4a2(t1-t2)2]
=√[a29X25/16 + 4a2X25/4]
=√(625a2/16)
=25a/4 units. (should be the ans.....)
{y i have not taken t1=-1/t2 in the very beginning is becoz... then in the end terms like t^4 will come and things will look massive....:) }
