Find the length of the latus rectum of the ellipse 25x² + 4y² = 100 ?

CONCEPT:

We have two standard forms of the ellipse, i.e
i) x2/a2 + y2/b2 = 1 & ii)  x2/b2 + y2/a2 = 1

In both cases a > b and b2 = a2(1 – e2), e < 1.

In (i) major axis is along the x-axis and the minor along the y-axis and in (ii) major axis is along y-axis and minor along the x-axis

Here our Questions match with case (ii). So for case (ii) Length of the latus rectum = 2b2/a

The following are the properties of a vertical ellipse \(\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1\) where 0 < b < a

• Its centre is (0, 0)
• Its vertices are (0, - a) and (0, a)
• Its foci are (0, - ae) and (0, ae)
• Length of major axis is 2a
• Length of minor axis is 2b
• Equation of major axis is x = 0
• Equation of minor axis is y = 0
• Length of latus rectum is \(\frac{2b^2}{a}\)
• Eccentricity of ellipse is \(e = \frac{\sqrt {a^2-b^2}}{a}\)

CALCULATION:

Given: Equation of ellipse is 25x2 + 4y2 = 100

The given equation can be re-written as: \(\frac{{{x^2}}}{{{4}}} + \frac{{{y^2}}}{{{25}}} = 1\)

As we can see that, the given ellipse is a vertical ellipse.

So, by comparing the given equation of ellipse with \(\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1\)  we get,

⇒ a² = 25 and b² = 4

As we know that, length of latus rectum of an ellipse is given by \(\frac{2b^2}{a}\)

So, length of latus rectum for the given ellipse is: \(\frac{2\times4}{5} = \frac{8}{5}\) units