7.5 Conic sections  (Page 4/23)

Divide both sides by $a^2-c^2.$ This gives the equation

If we refer back to [link] , then the length of each of the two green line segments is equal to a . This is true because the sum of the distances from the point Q to the foci $F\text{and}{F}^{\prime }$ is equal to 2 a , and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length a and leg lengths b and c . From the Pythagorean theorem, $a^2+b^2=c^2$ and $b^2=a^2-c^2.$ Therefore the equation of the ellipse becomes

Finally, if the center of the ellipse is moved from the origin to a point $\left(h,k\right),$ we have the following standard form of an ellipse.

If the major axis is horizontal, then the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is called vertical. The equation of an ellipse is in general form if it is in the form $A{x}^2+B{y}^2+Cx+Dy+E=0,$ where A and B are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square.

Finding the standard form of an ellipse

Put the equation $9x^2+4y^2-36x+24y+36=0$ into standard form and graph the resulting ellipse.

First subtract 36 from both sides of the equation:

$9x^2+4y^2-36x+24y=\mathrm{-36}.$

Next group the x terms together and the y terms together, and factor out the common factor:

$\begin{array}{ccc}\hfill \left(9x^2-36x\right)+\left(4y^2+24y\right)& =\hfill & \mathrm{-36}\hfill \\ \hfill 9\left(x^2-4x\right)+4\left(y^2+6y\right)& =\hfill & \mathrm{-36.}\hfill \end{array}$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives ${\left(\frac{\mathrm{-4}}{2}\right)}^2=4.$ In the second set of parentheses, take half the coefficient of y and square it. This gives ${\left(\frac{6}{2}\right)}^2=9.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

$\begin{array}{}\\ 9\left(x^2-4x+4\right)+4\left(y^2+6y+9\right)=\mathrm{-36}+36+36\hfill \\ 9\left(x^2-4x+4\right)+4\left(y^2+6y+9\right)=36.\hfill \end{array}$

Now factor both sets of parentheses and divide by 36:

$\begin{array}{}\\ \hfill 9{\left(x-2\right)}^2+4{\left(y+3\right)}^2& =\hfill & 36\hfill \\ \hfill \frac{9{\left(x-2\right)}^2}{36}+\frac{4{\left(y+3\right)}^2}{36}& =\hfill & 1\hfill \\ \hfill \frac{{\left(x-2\right)}^2}{4}+\frac{{\left(y+3\right)}^2}{9}& =\hfill & 1.\hfill \end{array}$

The equation is now in standard form. Comparing this to [link] gives $h=2,$ $k=\mathrm{-3},$ $a=3,$ and $b=2.$ This is a vertical ellipse with center at $\left(2,\mathrm{-3}\right),$ major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

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According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in [link] (a). Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth.