Different Types of Parabola Equations

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).parabola

(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.

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