Chapter 9.2, Problem 37E

(a)

To determine

To find:The standard form of the equation of ellipse $49x^2+4y^2-196=0$ .

Answer to Problem 37E

The standard form of the equation of ellipse $49x^2+4y^2-196=0$ is $\frac{x^2}{2^2}+\frac{y^2}{7^2}=1$ .

Explanation of Solution

Given information:

The equation of the ellipse is $49x^2+4y^2-196=0$ .

Calculation:

Rewrite the given equation of ellipse.

  $\begin{array}{c}49x^2+4y^2-196=0\\ 49x^2+4y^2=196\\ \frac{49x^2}{196}+\frac{4y^2}{196}=1\\ \frac{x^2}{4}+\frac{y^2}{49}=1\\ \frac{x^2}{2^2}+\frac{y^2}{7^2}=1\end{array}$

Therefore, the standard form of the equation of ellipse $49x^2+4y^2-196=0$ is $\frac{x^2}{2^2}+\frac{y^2}{7^2}=1$ .

(b)

To determine

To find:The center, vertices, foci and eccentricity of the ellipse $49x^2+4y^2-196=0$ .

Answer to Problem 37E

The center, vertices, foci and eccentricity of the ellipse $49x^2+4y^2-196=0$ are $\left(0,0\right)$ , $\left(0,\pm 7\right)$ , $\left(0,\pm 3\sqrt{5}\right)$ and $0.96$ respectively.

Explanation of Solution

Given information:

The equation of the ellipse is $49x^2+4y^2-196=0$ .

Calculation:

As calculated in part(a), the standard form of the equation of ellipse is $\frac{x^2}{2^2}+\frac{y^2}{7^2}=1$ .

Compare the equation of ellipse $\frac{x^2}{2^2}+\frac{y^2}{7^2}=1$ with general equation $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$ .

This gives $a=7$ , $b=2$ , $h=0$ and $k=0$ . The major axis of the ellipse is vertical.

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ c^2=49-4\\ c^2=45\\ c=3\sqrt{5}\end{array}$

Use the formula for eccentricity.

  $\begin{array}{c}e=\frac{c}{a}\\ =\frac{3\sqrt{5}}{7}\\ \approx 0.96\end{array}$

The center of the ellipse is $\left(h,k\right)=\left(0,0\right)$ .

The vertices of the ellipse are:

  $\begin{array}{c}\left(h,k\pm a\right)=\left(0,0\pm 7\right)\\ =\left(0,\pm 7\right)\end{array}$

The foci of the ellipse are:

  $\begin{array}{c}\left(h,k\pm c\right)=\left(0,0\pm 3\sqrt{5}\right)\\ =\left(0,\pm 3\sqrt{5}\right)\end{array}$

The eccentricity of ellipse is $e=0.96$ .

Therefore, The center, vertices, foci and eccentricity of the ellipse $49x^2+4y^2-196=0$ are $\left(0,0\right)$ , $\left(0,\pm 7\right)$ , $\left(0,\pm 3\sqrt{5}\right)$ and $0.96$ respectively.

(c)

To determine

To graph:The ellipse of equation $49x^2+4y^2-196=0$ .

Answer to Problem 37E

The graph of the ellipse $49x^2+4y^2-196=0$ is shown in Figure(1).

Explanation of Solution

Given information:

The equation of the ellipse is $49x^2+4y^2-196=0$ .

Calculation:

The graph of the ellipse $49x^2+4y^2-196=0$ is shown below.

  

Figure(1)

Verify the result graphically:

The graph of the equation $49x^2+4y^2-196=0$ with the help of graphing utility.

  

Figure(2)

As both the graphs gives the same values at each point.

Therefore, both the graphs are similar.