Question

The centre of the ellipse 8x^2+ 6y^2-16x +12y + 13 = 0 is at
(a)(1,1)
(b)(-1,1)
(c)(1,-1)
(d) none of these​

Answer 1

The answer is (b) (-1,1).

Step-by-step explanation:

Given:

The ellipse equation is

$\rightarrow \: 8 {x}^{2} + 6 {y}^{2} - 16x + 12y + 13 = 0 \\ \rightarrow \:8 {x}^{2} - 16x + 8 + 6{y}^{2} + 12y + 6 - 1 = 0 \\ \rightarrow \:8( {x}^{2} - 2x + 1) + 6( {y}^{2} + 2y + 1) = 1 \\ \rightarrow \:8 {(x - 1)}^{2} + 6 {(y + 1)}^{2} = 1$

The center of this ellipse is (-1,1).

Answer 2

The correct option is C.

Step-by-step explanation:

The given equation is

$8x^2+ 6y^2-16x +12y + 13 = 0$

It can be rewritten as

$8x^2+ 6y^2-16x +12y +8+6-1( = 0$

$(8x^2-16x+8)+( 6y^2+12y +6) -1=0$

$8(x^2-2x+1)+6(y^2+2y+1) -1=0$

$8(x-1)^2+6(y+1)^2 -1=0$

Add 1 on both sides.

$8(x-1)^2+6(y+1)^2=1$           .....(1)

The standard form of an ellipse is

$\dfrac{(x-h)^2}{a}+\dfrac{(y-k)^2}{b}=1$        .... (2)

where, (h,k) is center.

On comparing (1) and (2) we get

$h=1,k=-1$

The center of the ellipse is at (1,-1).

Therefore, the correct option is C.

#Learn more

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