Length of Latus Rectum of Parabola Formula

Here you will learn formula to find the length of latus rectum of parabola with examples.

Let’s begin –

Latus Rectum of Parabola

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

In the given figure, LSL’ is the latus rectum of the parabola \(y^2\) = 4ax.parabola

By the symmetry of the curve SL = SL’ = \(\lambda\) (say). So, the coordinates of L are \((a, \lambda)\).

Since L lies on \(y^2\) = 4ax, therefore

\({\lambda}^2\) = \(4a^2\)  \(\implies\)  \(\lambda\) = 2a

\(\implies\)  LL’ = \(2\lambda\) = 4a

Hence, Latus Rectum = 4a

Note : The length of latus rectum of all other forms of parabola i.e. \(x^2\) = 4ay , \(y^2\) = -4ax and \(x^2\) = -4ay is also equal to 4a.

Also Read : Different Types of Parabola Equations

Coordinates of Latus Rectum

The coordinates of L and L’ , end points of the latus rectum, are (a, 2a) and (a, -2a) respectively.

Example : For the given parabola, find the length of the latus rectum:

(i) \(y^2\) = 8x

(ii) \(x^2\) = -16y

Solution :

(i) The given parabola is of the form \(y^2\) = 4ax, where 4a = 8 i.e. a = 2.

Hence, Length of latus rectum = 4a = 8

(ii) The given parabola is of the form \(x^2\) = -4ay, where 4a = 16 i.e. a = 4.

Hence, Length of latus rectum = 4a = 16

Example : Find the latus rectum of the parabola \(y^2 – 8y – x + 19\) = 0

Solution : The given equation is

\(y^2 – 8y – x + 19\) = 0  \(\implies\)  \(y^2 – 8y\) = x – 19

\(\implies\)  \(y^2 – 8y + 16\) = x – 19 + 16

\(\implies\)  \((y – 4)^2\) = (x – 3)

The equation is of the form \(y^2\) = 4ax, where 4a = 1 i.e. a = 1/4.

Hence, Length of Latus Rectum is 4a = 1.

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