Question
The equation $$\left| {z - i} \right| + \left| {z + i} \right| = k,k > 0,$$ can represent an ellipse if $$k$$ is
A.
$$1$$
B.
$$2$$
C.
$$4$$
D.
None of these
Answer :
$$4$$
Solution :
$$\eqalign{
& \sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} + \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}} = k\,\,\,\,\,.....\left( 1 \right) \cr
& {\text{or, }}{x^2} + {\left( {y - 1} \right)^2} - {x^2} - {\left( {y + 1} \right)^2} = k\left\{ {\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} - \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}} } \right\} \cr
& \therefore \,\,\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} - \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}} = \frac{{ - 4y}}{k}\,\,\,\,\,.....\left( 2 \right) \cr} $$
From (1) and (2),
$$\eqalign{
& 2\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} = k - \frac{{4y}}{k} \cr
& \Rightarrow \,\,4{x^2} + \left( {4 - \frac{{16}}{{{k^2}}}} \right){y^2} = {k^2} - 4. \cr} $$
For an ellipse, $$4 - \frac{{16}}{{{k^2}}} > 0,{k^2} - 4 > 0.$$
For $$k = 1, 2$$ the co-efficient of $${y^2}$$ is negative or zero.