Question

find the length of major and minor axis 4x^2+3y^2=12​

Answer 1

Answer:

4x²/12 + 3y²/12 = 12/12

x²/3 + y²/4 = 1

compare with x²/a² + y²/b² = 1

a² = 3 => a = ±√3

b² = 4 => b = ±2

then majaor axis = 2b = 2x2 = 4 unit

minor axis = 2a = 2x√3 = 2√3 unit

Answer 2

Answer:

Length of the major axis  =  4 units

Length of the minor axis  = 2√3units

Step-by-step explanation:

Given the equation of the ellipse 4x² +3y² = 12

Recall the concept  $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ---------(1)

is the equation of the ellipse,

The major axis = 2bThe length of t

The length of the minor axis  = 2a

Solution

To find the length of the major axis and minor axis, Convert the given equation is the form  is  $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Given equation is 4x² +3y² = 12

Dividing throughout by 12

$\frac{4x^2}{12} +\frac{3y^2}{12} = 1$

$\frac{x^2}{3} +\frac{y^2}{4} = 1$

$\frac{x^2}{(\sqrt{3}^2) } +\frac{y^2}{2^2} = 1$

Comparing with equation (1) we get

a = √3 and b  = 2

2a = 2√3 and 2b = 4

Length of the major axis  =  2b = 4 units

Length of the minor axis  = 2a = 2√3units

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