The co-ordinates of the foci of the ellipse 3 | vedclass

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Question

(c) $3{x^2} - 12x + 4{y^2} - 8y = - 4$==>$3{(x - 2)^2} + 4{(y - 1)^2} = 12$

==> $\frac{{{{(x - 2)}^2}}}{4} + \frac{{{{(y - 1)}^2}}}{3} = 1$

Vedclass · Accepted Answer

$(1, 1), (3, 1)$

Vedclass · Answer

(c) $3{x^2} - 12x + 4{y^2} - 8y = - 4$==>$3{(x - 2)^2} + 4{(y - 1)^2} = 12$

==> $\frac{{{{(x - 2)}^2}}}{4} + \frac{{{{(y - 1)}^2}}}{3} = 1$

==> $\frac{{{X^2}}}{4} + \frac{{{Y^2}}}{3} = 1$

 $e = \sqrt {1 - \frac{3}{4}} = \frac{1}{2}$. 

Foci are $\left( {X = \pm 2 	imes \frac{1}{2},\,Y = 0} ight)$

$i.e.$, $(x - 2 = \pm 1,\,\,y - 1 = 0)$$ = (3,\,1)$ and $(1,\,1)$.

The co-ordinates of the foci of the ellipse $3{x^2} + 4{y^2} - 12x - 8y + 4 = 0$ are

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