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 1

Hint: In this problem, we have to find the value of k if it represents an ellipse. We can first take \[z=x+iy\] and substitute in the given equation. We can then square them and get an equation which represents an ellipse and we can write it by the given condition and solve for k, to get the required value of the term k.

Complete step by step solution:
We know that the equation given is,
 \[\Rightarrow \left| z-i \right|+\left| z+i \right|=k\] ……. (1)
Where \[k>0\].
We can now assume \[z=x+iy\].
We can now substitute \[z=x+iy\] in (1), we get
\[\Rightarrow \left| x+iy-i \right|+\left| x+iy+i \right|=k\]
We can now write the above step as,
\[\Rightarrow \left| x+i\left( y-1 \right) \right|+\left| x+i\left( y+1 \right) \right|=k\]
We can now take modulus, we get
\[\Rightarrow \sqrt{{{x}^{2}}+{{\left( y-1 \right)}^{2}}}+\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}=k\]
We can now write the above step as,
\[\Rightarrow \sqrt{{{x}^{2}}+{{\left( y-1 \right)}^{2}}}=k-\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}\]
We can now square on both sides, we get
\[\Rightarrow {{\left( \sqrt{{{x}^{2}}+{{\left( y-1 \right)}^{2}}} \right)}^{2}}={{\left( k-\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}} \right)}^{2}}\]
We can now simplify the above terms using the whole square expansion formula, we get
\[\Rightarrow {{x}^{2}}+{{\left( y-1 \right)}^{2}}={{k}^{2}}+{{x}^{2}}+{{\left( y+1 \right)}^{2}}-2k\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}\]
We can simplify the above step by cancelling the similar terms, we get
\[\Rightarrow 2k\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}={{k}^{2}}+{{\left( y+1 \right)}^{2}}-{{\left( y-1 \right)}^{2}}\]
We can now expand and simplify the RHS of the above step,
\[\begin{align}
  & \Rightarrow 2k\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}={{k}^{2}}+{{y}^{2}}+1+2y-{{y}^{2}}-1+2y \\
 & \Rightarrow 2k\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}={{k}^{2}}+4y \\
\end{align}\]
We can again take squaring on both sides in the above step,
\[\Rightarrow 4{{k}^{2}}\left( {{x}^{2}}+{{\left( y+1 \right)}^{2}} \right)={{k}^{4}}+16{{y}^{2}}+8{{k}^{2}}y\]
We can now expand the LHS and simplify the step, we get
\[\begin{align}
  & \Rightarrow 4{{k}^{2}}\left( {{x}^{2}}+{{y}^{2}}+1+2y \right)={{k}^{4}}+16{{y}^{2}}+8{{k}^{2}}y \\
 & \Rightarrow \left( 4{{k}^{2}}{{x}^{2}}+4{{k}^{2}}{{y}^{2}}+4{{k}^{2}}+8{{k}^{2}}y \right)={{k}^{4}}+16{{y}^{2}}+8{{k}^{2}}y \\
\end{align}\]
We can now simplify the above step, we get
\[\Rightarrow 4{{k}^{2}}{{x}^{2}}+{{y}^{2}}\left( 4{{k}^{2}}-16 \right)+4{{k}^{2}}-{{k}^{4}}=0\]
Therefore, we can see that it represents an ellipse if
\[4{{k}^{2}}-16>0\] (by the given condition).
\[\begin{align}
  & \Rightarrow 4{{k}^{2}}>16 \\
 & \Rightarrow {{k}^{2}}>4 \\
 & \Rightarrow k>2 \\
\end{align}\]
We have the value of k greater than 2 such that we have 3, 4 and here we are given 4 as the option.
Therefore, the answer is option C. 4.

Note: We should always remember that the algebraic whole square formula is expanded as \[{{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy\]. We should know to simplify the terms step by step and if we square take square root, then it will be cancelled as the radical means half to the power.