Here you will learn what is the formula for the length of latus rectum of ellipse with examples..
Letโs begin โ
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)ย \(16x^2 + 25y^2\) = 400
(ii)ย \(x^2 + 4y^2 โ 2x\) = 0
Solution :
(i)ย 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) = \(2b^2\over a\)
\(\implies\)ย L = \(2\times 16\over 5\) = \(32\over 5\)
(ii) We have,
\(x^2 + 4y^2 โ 2x\) = 0
\(\implies\) \((x โ 1)^2\) + 4\((y โ 0)^2\) = 1
\(\implies\)ย \((x โ 1)^2\over 1^2\) + \((y โ 0)^2\over (1/2)^2\) = 1
Here, a = 1 and b = 1/2
Clearly a > b,
Therefore, Length of the Latus Rectum (L) = \(2b^2\over a\)
\(\implies\)ย L = \(2\times 1/4\over 1\) = \(1\over 2\)
Note : For the ellipse \((x โ h)^2\over a^2\) + \((y โ k)^2\over b^2\) = 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 = \(\pm be + k\).
