91Ó°ÊÓ

The given equation of an ellipse $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$where $AandB$are non-zero constants.

Any conic section can be defined as the locus of points with constant distances to a point and a line. That ratio is known as eccentricity, and it is commonly represented by the symbol$e$e .

If $A>B$

The eccentricity of an ellipse is defined as

$e=\frac{\sqrt{A^2-B^2}}{A}$

Any conic section can be defined as the locus of points whose distances to a point and a line are in a constant ratio. That ratio is called eccentricity, commonly denoted by $e$

If $A<B$

The eccentricity of an ellipse is defined by

$e=\frac{\sqrt{B^2-A^2}}{B}$

When the eccentricity of an ellipse is zero, it becomes a circle. If the eccentricity is one, the segment is a straight line.

If $A>B$

The eccentricity is given by $e=\frac{\sqrt{A^2-B^2}}{A}$

Then, role="math" localid="1654123831261" $$\begin{gathered} \underset{A→B}{\lim }\frac{\sqrt{A^2-B^2}}{A}=\frac{\sqrt{B^2-B^2}}{B} \\ =\frac{0}{B} \end{gathered}$$

$\underset{A→B}{\lim }\frac{\sqrt{A^2-B^2}}{A}=0$

The ellipse becomes more circular.

If $A>B$

The eccentricity is given by $e=\frac{\sqrt{A^2-B^2}}{A}$

Then,

$$\begin{gathered} \underset{A→∞}{\lim }\frac{\sqrt{A^2-B^2}}{A}=\frac{\sqrt{{∞}^2-B^2}}{∞} \\ =\frac{∞}{∞} \end{gathered}$$

$=1$

The ellipse elongates and flattens.

Hence, the answer is$\underset{A→B}{\lim }\frac{\sqrt{A^2-B^2}}{A}=0$that elongates and flattens