Here, you will learn Different Types of Parabola and Standard equations of parabola, focal chord, double ordinate and latus rectum of parabola.

Let’s begin –

## What is Parabola ?

A parabola is the locus of a point which moves in a plane, such that its distance from a fixed point(focus) is equal to its perpendicular distance from a fixed straight line(directrix).

The Standard equation of parabola is **\(y^2 = 4ax\) **and it is shown in figure. For this parabola :

(i) **Vertex** is (0,0).

(ii) **focus** is (a,0)

(iii) **Axis** is y = 0

(iv) **Directrix** is x + a = 0

**(a) Focal distance :**

The distance of a point on the parabola from the focus is called the focal distance of the point.

**(b) Focal chord :**

A chord of the parabola, which passes through the focus is called a focal chord.

**(c) Double ordinate :**

A chord of the parabola perpendicular to the axis of the symmetry is called double ordinate.

**(d) Latus rectum :**

A double ordinate passing through the focus or a focal chord perpendicular to the axis of parabola is called latus rectum.

For \(y^2 = 4ax\).

Length of the latus rectum = 4a

Length of the semi latus rectum = 2a

Ends of the latus rectum are L(a, 2a) & L'(a, -2a).

**Note :**

*(i) Perpendicular distance from focus on the directrix = half the latus rectum.*

*(ii) Vertex is middle point of the focus & point of intersection of directrix & axis.*

*(iii) Two parabolas are said to be equal if they have the same latus rectum.*

## Different Types of Parabola & Standard Equations of Parabola

Four different types of parabola equations are

\(y^2\) = 4ax ; \(y^2\) = -4ax ; \(x^2\) = 4ay ; \(x^2\) = -4ay.

One I had shown above and three others are shown below.

**\(y^2\) = -4ax**

**\(x^2\) = 4ay**

**\(x^2\) = -4ay**

Parabola | Vertex | Focus | Axis | Directrix |
---|---|---|---|---|

\(y^2\) = 4ax | (0,0) | (a,0) | y = 0 | x = -a |

\(y^2\) = -4ax | (0,0) | (-a,0) | y = 0 | x = a |

\(x^2\) = +4ay | (0,0) | (0,a) | x = 0 | y = -a |

\(x^2\) = -4ay | (0,0) | (0,-a) | x = 0 | y = a |

\((y-k)^2\) = 4a(x-h) | (h,k) | (h+k,k) | y = k | x+a-h = 0 |

\((x-p)^2\) = 4b(y-q) | (p,q) | (p,b+q) | x = p | y+b-q = 0 |

Length of Latus rectum | Ends of Latus rectum | Parametric equation | Focal length |
---|---|---|---|

4a | (a,\(\pm\)2a) | (a\(t^2\), 2at) | x + a |

4a | (-a,\(\pm\)2a) | (-a\(t^2\), 2at) | x – a |

4a | (\(\pm\)2a,a) | (2at, a\(t^2\)) | y + a |

4a | (\(\pm\)2a,-a) | (2at, -a\(t^2\)) | y – a |

4a | (h+a, k\(\pm\)2a) | (h+a\(t^2\), k+2at) | x – h + a |

4b | (p\(\pm\)2a, q+a) | (p+2at, q+a\(t^2\)) | y – q + b |

Example : Find the vertex, axis, directrix, focus, latus rectum and the tangent at vertex for the parabola \(9y^2 – 16x – 12y – 57\) = 0.

Solution : The given equation can be written as \(({y-2\over 3})^2\) =
\(16\over 9\)\(({x + 61\over 16})\) which is of the form \(y^2\) = 4ax. Hence the vertex is (-\(61\over 16\), \(2\over 3\))

The axis is y – \(2\over 3\) = 0 \(\implies\) y = \(2\over 3\)

The directrix is x + a – h = 0 \(\implies\) x + \(61\over 16\) + \(4\over 9\) \(\implies\) x = \(-613\over 144\)

The focus is (h+a, k) \(\implies\) (\(-485\over 144\), \(2\over 3\))

Length of the latus rectum = 4a = \(16\over 9\)

The tangent at the vertex is x – h = 0 \(\implies\) x = \(-61\over 16\)

**Position of a point relative to a parabola :**

The point (\(x_1\),\(y_1\)) lies outside, on or inside the parabola \(y^2\) = 4a\(x_1\) is positive, zero or negative.