Question

The equation $\left| {z - i{\text{ }}} \right| + {\text{ }}\left| {z + i} \right|{\text{ }} = {\text{ }}k$ represents an ellipse if $k$ equals:
A. $1$
B. $2$
C. $4$
D. $-1$

Answer 1

Hint: The location of points in a plane whose sum of separations from two fixed points is a constant value is known as an ellipse. The conic section, which is the intersection of a cone with a plane that does not intersect the base of the cone, includes the ellipse.
To determine the value of $k$, we need to consider the complex number and then take the modulus of the equation of the question. Then square both sides of the equation.

Formula Used:
$(a+b)^2=a^2 + b^2 + 2ab$
$(a-b)^2=a^2 + b^2 - 2ab$

Complete step by step Solution:
In the question, the equation $\left| {z - i{\text{ }}} \right| + {\text{ }}\left| {z + i} \right|{\text{ }} = {\text{ }}k$ is given that represents an ellipse,
Consider $z = x + iy$, and substitute in the given equation, then:
$\left| {(x + iy) - i{\text{ }}} \right| + {\text{ }}\left| {(x + iy) + i} \right|{\text{ }} = {\text{ }}k$
As we know that a modulus function gives the magnitude of a number, so taking modulus in the above equation:
$\surd ({x^2} + {\left( {y - 1} \right)^2}) + {\text{ }}\surd ({x^2} + {\left( {y + 1} \right)^2}){\text{ }} = {\text{ }}k$
Subtract both sides of the above equation from $(\surd ({x^2} + {\left( {y + 1} \right)^2}))$,
$\surd ({x^2} + {\left( {y - 1} \right)^2}) + {\text{ }}\surd ({x^2} + {\left( {y + 1} \right)^2}) - \surd ({x^2} + {\left( {y + 1} \right)^2}){\text{ }} = {\text{ }}k - \surd ({x^2} + {\left( {y + 1} \right)^2}) \\$
$\surd ({x^2} + {\left( {y - 1} \right)^2}) = {\text{ }}k - \surd ({x^2} + {\left( {y + 1} \right)^2}) \\$
Squaring both sides of the equation, then:
${(\surd ({x^2} + {\left( {y - 1} \right)^2}))^2}{\text{ }} = {\text{ (}}k - \surd ({x^2} + {\left( {y + 1} \right)^2}){)^2}$
Use the algebraic identity to expand the above equation${(a - b)^2} = {a^2} + {b^2} - 2ab$,
${x^2} + {\left( {y - 1} \right)^2}\; = {\text{ }}{k^2} + {x^2} + {\left( {y + 1} \right)^2} - 2k\surd ({x^2} + {\left( {y + 1} \right)^2} \\$
   $\Rightarrow 2k\surd ({x^2} + {\left( {y + 1} \right)^2}\; = {\text{ }}{k^2} + {\left( {y + 1} \right)^2} - {\left( {y - 1} \right)^2} \\$
   $\Rightarrow 2k\surd ({x^2} + {\left( {y + 1} \right)^2}\; = {\text{ }}{k^2} + {y^2} + 2y + 1 - {y^2} + 2y - 1 \\$
   $\Rightarrow 2k\surd ({x^2} + {\left( {y + 1} \right)^2}\; = {\text{ }}{k^2} + 4y \\$
Again, squaring both sides of the obtained equation:
$(2k\surd {({x^2} + {\left( {y + 1} \right)^2}\;)^2} = {\text{ (}}{k^2} + 4y{)^2}$
Use the algebraic identity to expand the above equation ${(a + b)^2} = {a^2} + {b^2} + 2ab$, then:
$4{k^2}({x^2} + {\left( {y + 1} \right)^2}\; = {\text{ }}{({k^2} + 4y)^2} \\$
 $\Rightarrow 4{k^2}{x^2} + 4{k^2}{y^2} + 8{k^2}y + 4{k^2}\; = {\text{ }}{k^4} + 16{y^2} + 8{k^2}y \\$
 $\Rightarrow 4{k^2}{x^2} + {y^2}(4{k^2} - 16) + 4{k^2} - {k^4}\; = {\text{ }}0 \\$$i$
As we notice that the obtained equation $4{k^2}{x^2} + {y^2}(4{k^2} - 16) + 4{k^2} - {k^4}\; = {\text{ }}0$ represents an ellipse if $4{k^2} - 16{\text{ }} > {\text{ }}0$, then we get:
$4{k^2}\; > 16 \\$
 $\Rightarrow {k^2}\; > {\text{ }}\dfrac{{16}}{4} \\$
 $\Rightarrow {k^2}\; > {\text{ }}4 \\$
  $\Rightarrow \left| k \right|{\text{ }} > 2{\text{ }} \\$
Therefore, the value of $k$ is $4$.

Hence, the correct option is C.

Note: We must remember the complex number that will help to expand the equation. And when we do the problems related to complex numbers you should take care of signs while equating equations. $x$ and $y$ are real numbers and $i$ is a symbol showing imaginary parts. The symbol $i$ is generally called an imaginary unit. In this form of a complex number, $x$ is known as the real part of the equation and $y$ is the imaginary part of the solution.