Introduction to Conic Sections (original) (raw)

Last Updated : 19 May, 2026

A conic section, also referred to just as a 'Conic'****,** is a curve obtained by intersecting a plane with a cone.

• These curves can be obtained by intersecting a plane with a double right circular cone.
• Imagine a cone being cut by a knife at different places, creating different types of curves.
• The four main Conic sections are: Circle, Ellipse, Parabola, and Hyperbola.

Each type of conic section has unique properties and equations, making them essential for understanding orbital mechanics, designing optical systems, and solving quadratic equations.

1. Circle

The circle is a conic section in which the locus of the point is always equidistant from the center of one point. If the plane cuts the conic section at right angles, i.e., β = 90°, then we get a circle****.**

• **The general equation of the circle is, (x - h) 2 + (y - k) 2 = r 2 .
• **Coordinates of Focus: The circle's focus is its center, (h, k)
• **Directrix: Not applicable, as circles do not have directrices.

2. Ellipse

If the plane cuts the conic section at an angle less than 90°, i.e., α < β < 90°, then we get an ellipse. We define the parabola as the locus of all the points where the sum of distances from two fixed points (foci) is always in contact.

**Standard Equation:

• Horizontal Major Axis: \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
• Vertical Major Axis: \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1

**Coordinates of Foci:

• Horizontal ellipse: (h \pm c, k)
• Vertical ellipse: (h, k \pm c); Where c = \sqrt{a^2 - b^2}

**Equations of Directrices:

• Horizontal ellipse: x = h \pm \frac{a^2}{c}
• Vertical ellipse: y = k \pm \frac{a^2}{c}

3. Parabola

If the plane cuts the conic section at an angle where α is equal to β, i.e., α = β, then we get a parabola. The parabola is the locus of a point that moves in such a way that its distance is always the same distance from a fixed point (called the focus) and a given Line (called the directrix).

**Standard Equation:

• Vertical Parabola: (x - h)^2 = 4p(y - k)
• Horizontal Parabola: (y - k)^2 = 4p(x - h)

**Coordinates of Focus:

• Vertical parabola: (h, k + p)
• Horizontal parabola: (h + p, k)

**Equations of Directrix:

• Vertical parabola: y=k−p
• Horizontal parabola: x=h−p

4. Hyperbola

If the plane cuts the conic section at an angle where β is less than α, i.e. β ϵ [0, a], then we get a hyperbola. We define the hyperbola as the locus of a point where the ratio of distance from a fixed point (focus) and a fixed line (directrix) is always constant.

**Standard Equation:

• Horizontal Transverse Axis:\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
• Vertical Transverse Axis: \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1

**Coordinates of Foci:

• Horizontal hyperbola: (h \pm c, k)
• Vertical hyperbola: (h, k \pm c); Where c = \sqrt{a^2 + b^2}

**Equations of Directrices:

• Horizontal hyperbola: x = h \pm \frac{a^2}{c}
• Vertical hyperbola: y = k \pm \frac{a^2}{c}

Conic Sections Equations

The standard equations of the conic section are added in the table below.

**Solved Examples

**Example 1: Find the equation of a circle that has a center of (0, 0) and a radius of 5.
**Solution:

We have studied the formula for the equation of the circle.

(x-h)2 + (y - k)2 = r2

We just need to plug in the values in the formula.

Here, h = 0, k = 0 and r = 5
(x - 0)2 + (y - 0)2 = 52
⇒x2 + y2 = 52
⇒x2 + y2 = 25

**Example 2: Find the equation of the circle with center (-4, 5) and radius 4.
**Solution:

The formula for the equation of the circle.

(x-h)2 + (y - k)2 = r2

We just need to plug in the values in the formula.

Here, h = -4, k = 5 and r = 4
(x - (-4))2 + (y - 5)2 = 52
⇒(x + 4)2 + (y - 5)2 = 25
⇒x2 + 16 + 8x + y2 + 25 - 10y = 25
⇒x2 + 8x + y2 -10y + 16= 0

**Example 3: The equation given below is an equation of a circle. Find the radius and the centre: x2 + 6x + y - 4y = 3

**Solution:

We are given the equation, now to find out the radius and the center. We need to rearrange the equation such that this equation can come in the form given below.
(x-h)2 + (y - k)2 = r2
x2 + 6x + y2 - 4y = 3
⇒ x2 + (2)(3)x + y2 - 2(2)y = 3

We can see that these equations can be separated into two perfect squares.
⇒ x2 + (2)(3)x + 9 – 9 + y2 - 2(2)y + 4 – 4 = 3
⇒ (x + 3)2 - 9 + (y - 2)2 - 4 = 3
⇒ (x + 3)2 + (y - 2)2 = 3 + 4 + 9
⇒ (x + 3)2 + (y - 2)2 = 16

⇒ (x + 3)2 + (y - 2)2 = 42

Now comparing this equation with the standard equation of the circle, we notice,

h = -3, k = 2 and radius = 4.

**Example 4: Find the equation of the circle, with centre (-h, -k) and radius \sqrt{h^2 + k^2}

**Solution:

The standard equation of the circle is given by,

(x-h)2 + (y - k)2 = r2

Here, we have h = -h and k = -k and radius =√{h2 + k2}

Putting these values into the equation

(x + h)2 + (y + k)2 = (√{h2 + k2})2
x2 + h2 + 2hx + y2 + k2 + 2ky = h2 + k2
x2 + y2 + 2hx + 2ky = 0

**Example 5: Let's say we are given a line x + y = 2 and a circle that passes through the points (2, -2) and (3, 4). It is also given that the center of the circle lies on the line. Find out the radius-2d centre of the circle.

**Solution:

Let's say the equation of the circle is, (x - h)2 + (y - k)2 = r2

Now we know that the center of the circle lies on the line x + y = 2. Since the center of the circle is (h, k), it should satisfy this line.
h + k = 2

Putting the value of h from this equation into the equation of the circle.
(x - (2 - k))2 + (y - k)2 = r2

Now we also know that the circle satisfies the points (2, -2) and (3, 4). Putting (2, -2) in the above equation.
(2-(2 - k))2 + (-2 - k)2 = r2
⇒ k2 + (k + 2)2 = r2
⇒ k2 + k2 + 4 + 4k = r2
⇒ 2k2 + 4 + 4k = r2 .....(1)

Putting the equation (3, 4) is,
(x-(2 - k))2 + (y - k)2 = r2
⇒ (3 - (2 - k))2 + (4 - k)2 = r2
⇒(1 - k)2 + (4 - k)2 = r2
⇒ k2 - 2k + 1 + 16 - 8k + k2 = r2
⇒ 2k2 - 10k + 17 = r2 ......(2)

Solving these equations we get, h = 0.7, k = 1.7 and r2 = 12.58

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