# Find the length of the minor axis of the ellipse 16x2 + y2 = 16 ?

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Find the length of the minor axis of the ellipse 16x2 + y2 = 16 ?
1. 8
2. 10
3. 12
4. None of these
5. 16

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Correct Answer - Option 4 : None of these

CONCEPT:

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 2b2/a
• Eccentricity of ellipse is $e = \frac{\sqrt {a^2-b^2}}{a}$

CALCULATION:

Given: Equation of ellipse is 16x2 + y2 = 16

The given equation of ellipse can be re-written as: $\frac{{{x^2}}}{{{1}}} + \frac{{{y^2}}}{{{16}}} = 1$

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

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

⇒ a2 = 16 and b2 = 1

As we know that, for the vertical ellipse the length of the minor axis is given by 2b

So, the length of the major axis for the given ellipse is 2 ⋅ 1 = 2 units

Hence,  option D is the correct answer.