উপবৃত্তের ফোকাস নির্ণয়

Consider the standard form for the equation of an ellipse whose axes are parallel to the coordinate axes: $$\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$$ This ellipse has center at $(x_0, y_0)$ and axes of lengths $2a$ and $2b$. For convenience, we will translate this ellipse so that its center is at the origin. Now, we have the ellipse: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Because the foci lie on the major axis, which is the longer axis, we have two cases to consider; they are perfectly symmetric, so we may assume that $a\ge b$. In this case, the foci will lie on the $x$-axis, so their coordinates will be $F_1=(c_1, 0)$ and $F_2=(c_2, 0)$, where $-a < c_2 \le c_1 < a$. We wish to calculate $c_1$ and $c_2$ in terms of $a$ and $b$.

First, note that $P_1=(a, 0)$ and $P_2=(-a, 0)$ are on the ellipse: $$P_1F_1 + P_1F_2 = (a - c_1) + (a - c_2) = 2a - (c_1 + c_2)$$ $$P_1F_1 + P_1F_2 = (c_1 - (-a)) + (c_2 - (-a)) = 2a + (c_1 + c_2)$$ By the definition of an ellipse, $PF_1 + PF_2$ is a constant for all points on the ellipse. Therefore, $$2a - (c_1 + c_2) = 2a + (c_1 + c_2)$$ $$c_1 + c_2 = 0$$ This means that the foci are symmetric with respect to the origin. Perhaps this was obvious due to symmetry, but it's good to prove it. However, now that we know that $c_1+c_2 = 0$, it follows that the constant sum of distances $PF_1 + PF_2$ must equal $2a$.

Far less obvious is how far away the foci should be from the origin. Now, we use the points $P_3 = (0, b)$ and $P_4 = (0, -b)$ (it does not really matter which point we use). Let us use $P_3$:

$$P_3F_1 + P_3F_2 = \sqrt{(0 - c_1)^2 + (b - 0)^2} + \sqrt{(0 - c_2)^2 + (b - 0)^2}$$ $$P_3F_1 + P_3F_2 = 2\sqrt{b^2+c_1^2}$$

We know that this constant sum must equal $2a$. Therefore, $c_1^2 = a^2 - b^2$. Since $c_1+c_2 = 0$ and $c_1\ge c_2$, $$c_1 = \sqrt{a^2 - b^2}$$ $$c_2 = -\sqrt{a^2 - b^2}$$

If $b>a$, we would instead obtain, through exactly the same process: $$c_1 = \sqrt{b^2 - a^2}$$ $$c_2 = -\sqrt{b^2 - a^2}$$

Now, we can simply translate the ellipse back to its original location. For the ellipse: $$\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$$ If $a>b$, the foci are located at: $$F_1 = (x_0 + \sqrt{a^2 - b^2}, y_0)$$ $$F_2 = (x_0 - \sqrt{a^2 - b^2}, y_0)$$ and if $a<b$, the foci are located at: $$F_1 = (x_0, y_0 + \sqrt{b^2 - a^2})$$ $$F_2 = (x_0, y_0 - \sqrt{b^2 - a^2})$$

(Video) The Foci of an Ellipse

by Anusha Rahman

In this video by Anusha Rahman, you will learn about the foci of an ellipse.

Summary

An ellipse is defined as a closed curve that resembled a flattened circle.

The foci of an ellipse are two special points that play a crucial role in defining the shape of the ellipse, often denoted as $F_1$ and $F_2$. The sum of the distances from any point on the ellipse to the two foci is constant.

The foci of an ellipse are two fixed points such that the sum of the distances from any point on the ellipse to the two foci is constant. This constant sum is equal to the length of the major axis of the ellipse.

Foci of an Ellipse

The foci of an ellipse are two fixed points located along the major axis, equidistant from the center (labeled in green below). These points play a crucial role in defining the shape of the ellipse, as the sum of the distances from any point on the ellipse to each focus is constant.

The distance from the center of the ellipse to each focus is called the focal length, which is less than the length of the semi-major axis. In essence, the foci help to characterize the ellipse's eccentricity, influencing its overall shape and how elongated it appears.

Now that we understand what a focus is, we can move on to graphing ellipses from foci.