4Ans given is
A C D E
But how ??
TMH says k is a positive real number ???
|z - 4| - |z - 5i| = k will represent hyperbola provided k will be
(A) 6
(B) 7.5
(C) 1
(D) -1
(E) -3
MULTIPLE ANS QS
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14 Answers
1well for ||z - z1| - |z - z2|| = k to be hyperbola
k<|z1 - z2|
wer z1 and z2 r focuses of hyperbola
so k<√16+25=√41=6.4
so ACDE correct
4Ya ure working is correct
But my doubt is
IS THERE ANY CONDITION SAYING K>O ??
TMH COMPLEX nOS Says so ??
1i guess take mod both sides of the eq |z - 4| - |z - 5i| = k
so it becomes ||z - 4| - |z - 5i|| =| k|
now for it to be hyperbola |k|<|z1 - z2|
|k|<√41
so -√41< k<√41
i guess i made a mistake in post#5
1sorry i dun hav tmh
so i m not getting wat u r trying to say [2]
i guess it is that if k is +ve and k≠|z1-z2| then it represents hyperbola
it doesnt at all mean k cant be negative
that cond is wen k is +ve
thats how i reason it....maybe i m rong
1according to TMH
|z-z1|-|z-z2|=k
represent a hyperbola if k is +ve real nos.
and k≠|z1-z2| and also z1and z2are two focus of hyperbolaso according to above statement A,B,C should be the answer.....
29K shud be a positive number if the eqn is given like this ||z-4| - |z-5i|| = k and k can be either positive or negative if the eqn is of the form |z-4| - |z-5i| = k
66The equation will represent a hyperbola if the constant k is less than the distance between (4,0) and (0,5) i.e. √41. Hence, A, C, D, E.