Length of Latus Rectum of Ellipse Formula

Here you will learn what is the formula for the length of latus rectum of ellipse with examples..

Let’s begin –

Length of Latus Rectum of Ellipse

(i) For the ellipse \(x^2\over a^2\) + \(y^2\over b^2\) = 1, a > b

Length of the Latus Rectum = \(2b^2\over a\)

Equation of latus rectum is x = \(\pm ae\).

(ii) For the ellipse \(x^2\over a^2\) + \(y^2\over b^2\) = 1, a < b

Length of the Latus Rectum = \(2a^2\over b\)

Equation of latus rectum is y = \(\pm be\).

Also Read : Different Types of Ellipse Equations and Graph

Example : For the given ellipses, find the length of latus rectum.

(i)  \(16x^2 + 25y^2\) = 400

(ii)  \(x^2 + 4y^2 – 2x\) = 0

Solution :

(i)  We have,

\(16x^2 + 25y^2\) = 400 \(\implies\) \(x^2\over 25\) + \(y^2\over 16\),

where \(a^2\) = 25 and \(b^2\) = 16 i.e. a = 5 and b = 4

Clearly a > b,

Therefore, Length of the Latus Rectum (L) = \(2b^2\over a\)

\(\implies\)  L = \(2\times 16\over 5\) = \(32\over 5\)

(ii) We have,

\(x^2 + 4y^2 – 2x\) = 0

\(\implies\) \((x – 1)^2\) + 4\((y – 0)^2\) = 1

\(\implies\)  \((x – 1)^2\over 1^2\) + \((y – 0)^2\over (1/2)^2\) = 1

Here, a = 1 and b = 1/2

Clearly a > b,

Therefore, Length of the Latus Rectum (L) = \(2b^2\over a\)

\(\implies\)  L = \(2\times 1/4\over 1\) = \(1\over 2\)

Note : For the ellipse \((x – h)^2\over a^2\) + \((y – k)^2\over b^2\) = 1 with center (h. k),

(i) For ellipse a > b,

The equation of latus rectum is x = \(\pm ae + h\).

(ii) For ellipse a < b,

The equation of latus rectum is y = \(\pm be + k\).

Leave a Comment

Your email address will not be published.