13 ???????
usually<<<<<<<<<< how cum.>>>>>>>.
consider a parabola....
we can usually draw three normals to it from an external point....
..FIND THE LOCUS OF ALL THOSE POINTS FROM WHICH ONLY TWO NORMALS CAN BE DRAWN ....
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10 Answers
1normals 3 and locus of just 2 .
i dono witch metod bu i tink takin normal eqns
and simultaneosly solvin any 2 three times and then takin commun nos which satisfy all of em . but if it satisfy all of em then parabola haz 3 normals . itz a more weird sum where u got it >>>>?
or u made urself haaaaaaa >>>?<<
1no yaaar!! i havenot made it myself....
i can give the soln... but its a nice question... i liked it so i posted....
jus try a few more times.... its much of algebra rather than parabola :)
1no yaaar!! i havenot made it myself....
i can give the soln... but its a nice question... i liked it so i posted....
jus try a few more times.... its much of algebra rather than parabola :)
1okok i jus coudnt think how 2 go ahead .>>>
it willlll b gud if ya posted>>>>>>>>>>><<<<<<<
62Try this...
there will be one repeated root
so take the roots as t,t,k
now proceed.
1yes i tried lik this only naaaa
but i donno if anythins misssin it is somwher goin wrong
hav u got nishant >>>>>>>?
62yes dear..
take the roots of am2+2am-mh-k=0 as t,p,p
p+p+t=0
p.p+p.t+p.t=2-h/a
p.p.t=k/a
now eliminate t (substituing t=-2p in equation 2 and 3)
u get p2=(h/3a-2/3)
and p3=-k/a
so then eliminate p again.
(h/3a-2/3)3/2=-k/a
replace h,k by x,y