Question

The equation $\left|z-i\right|+\left|z+i\right|=k$ represents an ellipse if $k$ equals

• A. $1$
• B. $2$
• C. $4$
• D. $-1$

Answer 1

The correct option is C

$4$

Calculate the value of $k$:

The given equation of an ellipse is $\left|z-i\right|+\left|z+i\right|=k$.

Let $z=x+iy$.

So, $\left|x+iy-i\right|+\left|x+iy+i\right|=k$.

Taking modulus on both sides of the equation,

$\begin{array}{rcl}\sqrt{x^2+{\left(y-1\right)}^2}+\sqrt{x^2+{\left(y+1\right)}^2}& =& k\\ & \Rightarrow & \sqrt{x^2+{\left(y-1\right)}^2}=k-\sqrt{x^2+{\left(y+1\right)}^2}\end{array}$

Squaring both sides,

$\begin{array}{rcl}{x}^2+{\left(y-1\right)}^2& =& k^2+x^2+{\left(y+1\right)}^2-2k\sqrt{x^2+{\left(y+1\right)}^2}\\ & \Rightarrow & 2k\sqrt{x^2+{\left(y+1\right)}^2}=k^2+{\left(y+1\right)}^2-{\left(y-1\right)}^2\\ & \Rightarrow & 2k\sqrt{x^2+{\left(y+1\right)}^2}=k^2+4y\end{array}$

Again on squaring,

$\begin{array}{rcl}4k^2\left(x^2+{\left(y+1\right)}^2\right)& =& {\left(k^2+4y\right)}^2\\ & \Rightarrow & 4k^2{x}^2+4k^2{y}^2+8k^{2}y+4k^2=k^4+16y^2+8k^{2}y\\ & \Rightarrow & 4k^2{x}^2+y^2\left(4k^2-16\right)+4k^2-k^4=0\end{array}$

Above equation represents an ellipse, if $4k^2-16>0$.

$$\begin{gathered} k^2>\frac{16}{4} \\ {k}^2>4 \\ \Rightarrow k\in (-\infty ,-2)\cup (2,\infty ) \end{gathered}$$

Hence, the value of $k$ is $4$ and therefore option $\left(C\right)$ is correct.