Let us start with the conics’ introduction of circles, eclipses, parabolas, and hyperbolas which includes the set of curves formed by the intersection of the plane and double-napped right cone. In case if you are interested, then there are four curves which can be formed, and all are useful in the applications of math and science.

In the section of conics, we will see every type of curve and how to recognize it and graph it. Then we’ll come up with some common applications.

The section of the conic section is the curve which is obtained as the intersection of the cone surface with the plane; the three types are: eclipse, parabola, and hyperbolas.

The conic section can be drawn on the coordinate plane. Each section of conic has some of the features which includes at least one directrix and one focus. Parabola has one focus and directrix whereas eclipses and hyperbolas have two of each of these.

The conic section is the set of points whose focus distance is the constant multiple to the PP distance to the directrix of the conic section.

Some of the terms are defined below –

- Vertex – it is an extreme point on the conic section.
- Locus – is the set of all the points whose coordinates get satisfied with a given equation or situation.
- Focus – is the point which is used for constructing and defining the conic section at which the rays get reflected from the curve coverage.
- Asymptote – is the straight line which a curve approaches closely as it goes to infinite distance.
- Nappe – is one-half section of the double cone.
- Directrix – is a line which is useful for construction and defining the conic section. Parabola has only one directrix, whereas eclipse and hyperbolas have two of them.

**Every different section of conic in detail –**

We will go with eclipse, parabola, and hyperbola in detail as these three conic sections with foci and directrix, are labeled. Every type of conic section is discussed in depth below –

## Parabola

The parabola is the set of all the points whose distance is known as the fixed point, called focus. The focus is the same as the distance of the fixed line, known as directrix. The point half in-between the directrix and focus is known as the vertex of the parabola. Below is the figure of the parabola which is shown opening up and down. It also appears in the right and left the plane.

The parabola is written in two forms: standard form and vertex form.

y = ax^2 + bx + c (standard form)

- a, b, and c are coefficients.
- If a is positive, the parabola opens upward. If a is negative, the parabola opens downward.

y = a(x-h) ^2 + k (vertex form)

- (h,k) = vertex
- expand vertex form to turn into standard form
- x – coordinate of vertex = -b/2a

**Example 1 – The equation of a parabola is y=x^2+2x+10. Write the equation in vertex form.**

1. Find a.

Looking at the equation, a = 1.

2. Find vertex (h,k).

The x-coordinate of the vertex is -b/2a. Looking at the equation, b = 2, and a = 1 as determined in the previous step.

x = -2/2(1) = -2/2 = -1

Plug in x= -1 into the equation to get the y-coordinate of the vertex.

y = (-1) ^2 + 2(-1) + 10 = 9

The vertex is (-1, 9)

3. Write the equation in vertex form.

Since a, h, and k are determined, the vertex form can be written:

a(x-h) ^2 + k = 1(x- -1)^2 + 9 = (x+1)^2 + 9.

**Example 2 – Find the vertex of the parabola: y = (x – 2)² + 4.**

1. Find h.

Since the vertex form is written as a(x-h) ^2 + k, h = 2

2. Find k.

Since the vertex form is written as a(x-h) ^2 + k, k =4

3. Write the vertex.

The vertex is written as (h, k) = (2, 4).

**Example 3 – What is the graph of the following parabola y = –(x+1)² + 1?**

1. Substitute x values to obtain y values.

Starting with x = -4, -3, -2, -1, 0, 1, and 2, you get y values of y = -8, -3, 0, 1, 0, -3, and -8.

2. Plot the points in the standard x-y coordinate plane.

(-4, -8)

(-3, -3)

(-2, 0)

(-1, 1)

(0, 0)

(1, -3)

(2, -8)

## Eclipse

An eclipse is the set of the points for which the distance addition, from the two fixed points, is the same. In case of any eclipse, there are two foci, the two directeces. Below is the figure of the eclipse as it appears on the coordinate plane. The sum of the distance from any of the point on eclipse to the focus is the same.

The equation is written in the following forms:

x^{2}/a^{2} + y^{2}/b^{2} = 1 (center at origin)

(x-h)^{2}/a^{2} + (y-k)^{2}/b^{2} = 1 (center at [h,k])

- x-intercepts at -a and a
- y-intercepts at -b and b

**Example 1 – Graph the ellipse x ^{2}/36 + y^{2}/4 = 1.**

1. Find the center.

Because it is written in the form x^{2}/a^{2} + y^{2}/b^{2} = 1, the center is at the origin (0,0).

2. Find the x and y intercepts.

Looking at the equation, a^2 = 36, so a = 6. The x intercepts are -6 and 6.

b^2 = 4, so b = 2. The y intercepts are -2 and 2.

3. Graph the ellipse.

**Example 2 – Graph the ellipse x ^{2}/ 9 + y^{2}/ 4 = 1.**

1. Find the center.

Because it is written in the form x^{2}/a^{2} + y^{2}/b^{2} = 1, the center is at the origin (0,0).

2. Find the x and y intercepts.

Looking at the equation, a^2 = 9, so a = 3. The x intercepts are -3 and 3.

b^2 = 4, so b = 2. The y intercepts are -2 and 2.

3. Graph the ellipse.

**Example 3 – Graph the ellipse (x+3) ^{2}/ 9 + (y-5)^{2}/ 3 = 1.**

1. Find the center.

Because the graph is written in the form (x-h)^{2}/a^{2} + (y-k)^{2}/b^{2} = 1, h = -3, and k = 5. The center is (-3, 5)

2. Find a and b.

Because the center is not at (0,0), there is no guarantee that there will be an x or y intercept. So, the end points will be graphed from the center using the values of a and b. Looking at the equation, a^2 = 9, so a = 3.

b^2 = 3, so b = √3.

3. Graph the ellipse.

## Hyperbola

A hyperbola is the point where the difference between such distance from any two fixed points, is the same (here points we are talking are foci). There are two directrix and foci in a hyperbola. Below is the figure of the hyperbola.

The hyperbola is written in the following form:

x^{2}/a^{2} – y^{2}/b^{2} = 1 (center at origin)

(x-h)^{2}/a^{2} – (y-k)^{2}/b^{2} = 1 (center at [h,k])

Equations of asymptotes

- y
_{1}= -bx/a - y
_{2}= bx/a

**Example 1 – Graph the hyperbola x ^{2}/ 9 – y^{2}/ 4 = 1.**

1. Find a and b

Looking at the equation, a^2 = 9, so a = 3. b^2 = 4, so b = 2.

2. Find the asymptotes

y_{1} = – bx/a = – 2x/3

y_{2} = bx/a = 2x/3

3. Graph the asymptotes on the coordinate plane.

4. Find the x intercepts.

to find the x intercept, set y = 0 and solve for x.

x^{2}/ 9 – 0^{2}/ 4 = 1.

x^{2}/ 9 = 1

Multiply by 9 on both sides.

~~9~~ * x^{2}/~~9~~ = 1* 9

x^{2}= 9

Take the square root of both sides.

x = +3 and -3.

The hyperbola crosses the x-axis at ±3 and approaches the asymptotes.