Question

Foci of the ellipse 2x² + 3y2 - 4x -12y +13 = 0 are .. ​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Ellipse is}$

$\mathsf{2x^2+3y^2-4x-12y+13=0}$

$\underline{\textbf{To find:}}$

$\textsf{Foci of the given ellipse}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{2x^2+3y^2-4x-12y+13=0}$

$\mathsf{2(x^2-2x)+3(y^2-4y)+13=0}$

$\mathsf{2(x^2-2x+1-1)+3(y^2-4y+4-4)+13=0}$

$\mathsf{2((x-1)^2-1)+3((y-2)^2-4)+13=0}$

$\mathsf{2(x-1)^2-2+3(y-2)^2-12+13=0}$

$\mathsf{2(x-1)^2+3(y-2)^2=1}$

$\mathsf{\dfrac{(x-1)^2}{\dfrac{1}{2}}+\dfrac{(y-2)^2}{\dfrac{1}{3}}=1}$

$\mathsf{Comparing\;this\;with\;\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1\;we\;get\;}$

$\mathsf{a^2=\dfrac{1}{2}\;and\;b^2=\dfrac{1}{3}}$

$\mathsf{ae=\sqrt{a^2-b^2}}$

$\mathsf{ae=\sqrt{\dfrac{1}{2}-\dfrac{1}{3}}}$

$\mathsf{ae=\sqrt{\dfrac{3-2}{6}}}$

$\mathsf{ae=\sqrt{\dfrac{1}{6}}}$

$\mathsf{ae=\dfrac{1}{\sqrt{6}}}$

$\therefore\mathsf{Foci\;are\;(\pm\,ae,0)}$

$\implies\mathsf{\left(\pm\dfrac{1}{\sqrt{6}},0\right)}$

$\underline{\textbf{Find more:}}$

Answer 2

Given  : 2x² + 3y2 - 4x -12y +13 = 0  

To Find : Foci of the ellipse

Solution:

2x² + 3y² - 4x -12y +13 = 0  

=> 2x² - 4x  + 3y²  - 12y + 13 = 0

=> 2(x²  - 2x + 1) - 2 + 3(y² - 4y + 4) -  12 + 13 = 0

=> 2(x - 1)² + 3(y - 2)² = 1

=> (x-1)²/(1/2)  + (y - 2)²/(1/3) =  1

Hence (x- h)²/a² + (y-k)²/b² = 1

h = 1 , k = 2

a² = 1/2

b²  = 1/3

c² = 1/2 - 1/3  = 1/6  => c = 1/√6

foci are ( h + c , k ) and ( h - c , k)

=> ( 1 + 1/√6 , 2) and ( 1 - 1/√6 , 2)

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