Chapter A5.2, Problem 13E

To determine

To sketch: the ellipse and find the standard form of equation, foci, eccentricity, and directrix

Answer to Problem 13E

Thestandard form of equation is $\frac{{(x-1)}^2}{4}+\frac{{(y-4)}^2}{9}=1$

  

Explanation of Solution

Given:

  $(1,1),(3,4),(1,7),$ and $(-1,4).$

Calculation:

End points of the two axes are (1,1) to (1,7) and the other is from (3,4) to (-1,4) .

The former axis is 6 units long and the later is 4 units long. Therefore $a=3$ and $b=2$ and themajor axis is vertical. The center is the point $C(1,4)$ and the ellipse is given by:

  $\frac{{(x-1)}^2}{4}+\frac{{(y-4)}^2}{9}=1$

Here calculate other parameters as:

  $c^2=a^2-b^2$

  $c=\sqrt{5}$

Therefore the foci of the ellipse can be written as $[F(1,4\pm \sqrt{5})]$

The eccentricity is

  $e=\frac{c}{a}=\frac{\sqrt{5}}{3}$

And the directrix is

  $y=4\pm \frac{3}{\left(\frac{\sqrt{5}}{3}\right)}$

  $y=4\pm \frac{9\sqrt{5}}{5}$

  

Conclusion:

The standard form of equation is $\frac{{(x-1)}^2}{4}+\frac{{(y-4)}^2}{9}=1$

Foci is $[F(1,4\pm \sqrt{5})]$

Directrix is $y=4\pm \frac{9\sqrt{5}}{5}$

Eccentricity is $\frac{\sqrt{5}}{3}$