Question #f11fb

Question

1092 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-31T18:34:58

$"Centre is "(1,1)," and, Eccentricity "e=1/2$ .

Shifting the Origin to the point $(1,1)$ , suppose that, old co-ords.

$(x,y)" becomes new ones "(X,Y)"$ .

Therefore, $x=X+1, y=Y+1.$

Substituting in the given eqn., it becomes, $8X^2+6Y^2=1, i.e.,$

$X^2/(1/8)+y^2/(1/6)=1$ .

Comparing with the Standard Eqn. of Ellipse $: X^2/a^2+Y^2/b^2=1$ ,

$a^2=1/8, b^2=1/6 :. b^2>a^2$ . Then, the Eccentricity $e$ is given by,

$a^2=b^2(1-e^2) :. e^2=1-a^2/b^2=1-(1/8)/(1/6)=1-6/8=1/4$

$:. e=1/2$

The Centre is $(X,Y)=(0,0) rArr (x-1,y-1)=(0,0), i.e.,$

$" Centre "(1,1)$ .

Enjoy Maths.!