Question

Find the centre, the length of axes, eccentricity, foci of the ellipse 3x2 + 4y2 -12x – 8y +4

=0​

Answer 1

SOLUTION

TO DETERMINE

• The centre

• The length of axes

• Eccentricity

• Foci

For the ellipse

$\sf{3 {x}^{2} + 4 {y}^{2} - 12x - 8y + 4 = 0}$

EVALUATION

Here the given equation of the ellipse is

$\sf{3 {x}^{2} + 4 {y}^{2} - 12x - 8y + 4 = 0}$

Which can be rewritten as below

$\sf{3 {x}^{2} + 4 {y}^{2} - 12x - 8y + 4 = 0}$

$\sf{ \implies \: 3( {x}^{2} - 4x + 4) + 4 ({y}^{2} - 2y + 1) = 12}$

$\sf{ \implies \: 3 {(x - 2)}^{2} + 4 {(y - 1)}^{2} = 12}$

$\displaystyle \sf{ \implies \: \frac{ {(x - 2)}^{2} }{4} + \frac{ {(y - 1)}^{2} }{3} = 1}$

$\displaystyle \sf{ \implies \: \frac{ {(x - 2)}^{2} }{ {(2)}^{2} } + \frac{ {(y - 1)}^{2} }{ {( \sqrt{3}) }^{2} } = 1}$

Comparing the above equation with

$\displaystyle \sf{ \frac{ {(x - \alpha )}^{2} }{ {a}^{2} } + \frac{ {(y - \beta )}^{2} }{ {b }^{2} } = 1}$

We get

$\sf{ \alpha = 2 \: , \: \beta = 1 \: , \: a = 2 \:, \: b = \sqrt{3} }$

FIGURE

For figure refer to the attachment

DETERMINATION OF CENTRE

The centre is

$\sf{( \alpha \: , \: \beta ) = (2,1)}$

In the figure it is denoted by C

DETERMINATION OF LENGTH OF AXES

Length of Major axis = 2a unit = 4 unit

In figure it is denoted by AA'

Length of Minor axis = 2b unit = 2√3 unit

In figure it is denoted by BB'

DETERMINATION OF ECCENTRICITY

The Eccentricity is given by

$\displaystyle \sf{e = \sqrt{1 - \frac{ {b}^{2} }{ {a}^{2} } } }$

$\displaystyle \sf{= \sqrt{1 - \frac{ 3 }{ 4 } } }$

$\displaystyle \sf{= \sqrt{ \frac{ 1 }{ 4 } } }$

$\displaystyle \sf{= \frac{1}{2} }$

DETERMINATION OF FOCI

The Foci are given by

$= \sf{( \alpha \pm \: ae \: ,\: \beta ) }$

$\displaystyle \sf{= \bigg(2 \pm \frac{2}{2} \: ,\: 1 \bigg) }$

$\displaystyle \sf{= \bigg(2 \pm 1 \: ,\: 1 \bigg) }$

$\displaystyle \sf{= \bigg(3 \: ,\: 1 \bigg) \: \: and \: \: \bigg(1 \: ,\: 1 \bigg) }$

In figure it is denoted by S(3,1) & S' (1,1)

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