Q. 9. - Calculus

Question

The graph of the equation $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$is an ellipse for any nonzero constants A and B.

(a) If A > B, what is the eccentricity of the ellipse?

(b) If A < B, what is the eccentricity of the ellipse?

(c) Explain why the eccentricity, e, of an ellipse is always

between 0 and 1.

(d) If A > B, what is $\underset{A\to B}{\lim } e$? What happens to the shape

of the ellipse as A → B?

(e) If A > B, what is $\underset{A\to \infty }{\lim }e$? What happens to the shape of the ellipse as A→∞?

Step-by-Step Solution

Answer

Part a) The answer is $e=\frac{\sqrt{A^2-B^2}}{A}$

Part b) The answer is $e=\frac{\sqrt{B^2-A^2}}{B}$

Part c) If the eccentricity is 1 then it would be a straight line segment.

Part d) The answer is $\underset{A\to B}{\lim }\frac{\sqrt{A^2-B^2}}{A}=0$

Part e) The answer is $\underset{A\to B}{\lim }\frac{\sqrt{A^2-B^2}}{A}=0$

Part (a) Step 1: If A &#62; B , the objective is to find out the eccentricity of the given ellipse

The given equation of an ellipse $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$where $A and B$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

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

Part (b) Step 1: If A &#60; B the objective is to find out the eccentricity of the given ellipse.

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

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

Part (c) Step 1: The objective is to explain why the eccentricity, e, of an ellipse is always between 0 and 1.

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

Part (d) Step 1: If A &#62; B &#160; then the objective is to find the value of lim A &#8594; B &#160; e and write the shape of the ellipse as A &#8594; B

If $A>B$

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

Then, $$\begin{gathered} \underset{A\to B}{\lim }\frac{\sqrt{A^2-B^2}}{A}=\frac{\sqrt{B^2-B^2}}{B} \\ =\frac{0}{B} \end{gathered}$$

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

The ellipse becomes more circular. 

Part (e) Step 1: If A &#62; B &#160; then the objective is to find the value of lim A &#8594; &#8734; e and write the shape of the ellipse as A &#8594; &#8734;

If $A>B$

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

Then,

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

$=1$

The ellipse elongates and flattens.

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