Foci of an Ellipse: Overview, Questions, Preparation

Before understanding the foci of an ellipse, we need to know about an ellipse. When a curved line forms a closed-loop, it is known as an ellipse. An ellipse is a circle but is slightly squeezed into an oval. A conic with eccentricity less than one is known as an ellipse.

Foci of an ellipse

Focus is the point that defines an ellipse, hyperbola, or parabola. The plural of focus is foci. The constant ratio is known as the eccentricity of an ellipse, which elongates the ellipse and is denoted by ‘e’.

The formula for the calculation of foci of an ellipse is 
F = √(j² - ⁿ²⁾
where F = distance between the center of an ellipse and the foci

 j = semi-major axis

n = semi-minor axis

Let us look at some of the properties of an ellipse

Focus points (Foci) - Foci are two points on an ellipse.

Major and minor axis - The longest diameter of an ellipse is called the major axis, while an ellipse’s shortest diameter is called the minor axis. 

Centre- The midpoint inside the ellipse, which is the midpoint of the two intersecting foci. 

Tangent- A line that passes by touching at one point of an ellipse is known as a tangent.

Secant- A secant is a line that intersects an ellipse at two points.

In the chapter, ‘Conic sections,’ you would understand the sections of a cone, circle, parabola, ellipse, hyperbola, and foci related to it. These sections would be explained with proper examples in this chapter.

1. Suppose a parabolic reflector is 20 cm in diameter and 5 cm deep. Find the focus?
Solution. O_c = 5 cm
= y² = 4ax

= x - coordinate of point A = 5

= Ab = 20

= AC = 1/2 x AB = 10

= A (5, 10)

The point A lies on the parabola and it satisfies the equation y² = 4ax of the parabola.

= y² = 4 x 9 x 5

= 9 = 5

Focus = (0 ,10)

= (5, 0)

2. Find the coordinates of the foci and the ellipse’s vertices for the equation (x²/25) + (y²/9) = 1.
Solution.  Since the denominator of (x²/25) is greater than the denominator of (y²/9), the major axis is on the x-axis. 
So equating the equation with (x²/a²) + (y²/b²) = 1, we get a = 5 and b = 3
= c =  √(a² - b²)
= c = √(25 - 9)
= c = 4
So, the coordinates of the foci are (4,0) and (-4,0), the vertices are (5,0) and (-5,0).

3. Find the equation of an ellipse whose foci are (±5, 0) and vertices are (±13, 0)?
Solution. 
Since the vertices are on x-axis, so the equation would be
= (x²/a²) + (y²/b²) = 1

Given a = 13, c = ±5
= c² = a² - b²
= 25 = 169 - b²
= b = 12
So, the equation of the ellipse is (x2/169) + (y2/144) = 1

Q: What is an ellipse?

Q: What is the focus of an ellipse?

Q: What is the formula to calculate the foci of an ellipse?

Q: What is the major a minor axis of an ellipse?

Q: What is a secant of an ellipse?