Consider the given equation,
[tex]4(x-2)^2+5(y+2)^2=0[/tex]
Divide the equation by 20 i.e. LCM of 5 and 4,
[tex]\frac{(x-2)^2}{5}+\frac{(y+2)^2}{4}=0\Rightarrow\frac{(x-2)^2}{(\sqrt{5})^2}+\frac{(y+2)^2}{2^2}=0[/tex]
The equation does not resemble an ellipse or circle.
Either it is a point, or none of the above.
Let us check if the equation resembles a point.
Consider the equation as,
[tex]5(y+2)^2=-4(x-2)^2\Rightarrow(y+2)^2=-\frac{4}{5}(x-2)^2\Rightarrow y+2=\sqrt[]{-\frac{4}{5}(x-2)^2}[/tex]
Simplify the expression further,
[tex]y+2=(x-2).\sqrt[]{\frac{-4}{5}}\Rightarrow y=2+(x-2).\sqrt[]{\frac{-4}{5}}[/tex]
Now, observe that for every value of 'x' there exist some value of 'y'. It means that the equation corresponds to a point.
But note that there is a negative term in the square root i.e. imaginary term.
It means that this point must lie in a complex plane, not on the real plane.
Therefore, the answer to the given question is a point.nary
