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# 数学代写|微积分代写Calculus Assignment代考|Math1131Q Ellipses

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## 数学代写|微积分代写Calculus Assignment代考|Ellipses

Next we turn to the study of ellipses. An ellipse has not just one focus, but rather two foci.

Definition 3 ELLIPSE An ellipse is the set of points for which the sum of the distances to two fixed points (foci) is constant.

Just as with a circle and a parabola, the synthetic definition of ellipse does not reference coordinates or equations. A synthetic picture of an ellipse is in figure $9 .$

To draw an ellipse, place a thumb tack at each focus and tie the ends of a string on each thumb tack. Use a pencil to stretch the string tight. The tightened string should then look like one of the dashed brown paths in figure 9, and the pencil should be located at a point on the ellipse. Because the length of the string doesn’t change, wherever the pencil is located is a point on the ellipse. Keep drawing for a full rotation and the ellipse is complete.

To find an equation for an ellipse, we must introduce coordinates, as in figure 10. Place the foci at $(-c, 0)$ and $(c, 0)$, and let the constant length of the path be $2 a$. Then, for any point $(x, y)$ on the ellipse, the distance from $(x, y)$ to $(-c, 0)$ plus the distance from $(x, y)$ to $(c, 0)$ is $2 a$. Using the distance formula, we have

$$\sqrt{(x-(-c))^2+(y-0)^2}+\sqrt{(x-c)^2+(y-0)^2}=2 a .$$
Rearranging gives
$$\sqrt{(x+c)^2+y^2}=2 a-\sqrt{(x-c)^2+y^2} .$$
Squaring both sides and simplifying eventually yields
$$2 x c=4 a^2-4 a \sqrt{(x-c)^2+y^2}-2 x c,$$
which can be rearranged to
$$\sqrt{(x-c)^2+y^2}=a-\frac{c}{a} x .$$
Squaring both sides again, simplifying, rearranging, and factoring yields
$$\left(1-\frac{c^2}{a^2}\right) x^2+y^2=a^2-c^2,$$
which can be rearranged to
$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1 .$$

## 数学代写|微积分代写Calculus Assignment代考|Hyperbolas

Hyperbolas are defined in a manner similar to ellipses, but using the difference instead of the sum.

Definition 4 HYPERBOLA $A$ hyperbola is the set of points for which the difference of the distances to two fixed points (foci) is constant.
Once again we have a definition without reference to coordinates or equations. A synthetic picture of a hyperbola is in figure $17 .$

A hyperbola has two branches, one associated with each focus. The intersection of the focal axis with a branch of the hyperbola is a vertex of the hyperbola. To determine an equation for a hyperbola, we use the coordinates $(\pm c, 0)$ for the foci, which then places the center at $(0,0)$ and makes the focal axis horizontal. From any point $(x, y)$ on the hyperbola, we set the common difference in distances to the foci as $2 a$. The derivation is very much the same as for an ellipse, with the exception that since $c^2>a^2$, we set $b^2=c^2-a^2$, making $c=\sqrt{a^2+b^2}$.

## 数学代写|微积分代写Calculus Assignment代考|Ellipses

. . . . . . . . . . . .

$$\sqrt{(x-(-c))^2+(y-0)^2}+\sqrt{(x-c)^2+(y-0)^2}=2 a .$$

$$\sqrt{(x+c)^2+y^2}=2 a-\sqrt{(x-c)^2+y^2} .$$

$$2 x c=4 a^2-4 a \sqrt{(x-c)^2+y^2}-2 x c,$$

$$\sqrt{(x-c)^2+y^2}=a-\frac{c}{a} x .$$

$$\left(1-\frac{c^2}{a^2}\right) x^2+y^2=a^2-c^2,$$

$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1 .$$

## MATLAB代写

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