Question #a7bcf

1 Answer
Jan 28, 2018

The equation of ellipse is (x+y-1)^2/8+(y-x)^2/7 =1

Explanation:

Focii of the ellipse are at f1 (0,0) and f2 (1,1)

Distance between them f1f2= sqrt ((x_1-x_2)^2+(y_1-y_2)^2

or f1f2= sqrt((0-1)^2+(0-1)^2) =sqrt2

Distance of the focii from centre is c=sqrt2/2

Semi major axis length a=4/2=2 The relation of

c, a, b is c^2 = a^2 - b^2:. (sqrt2/2)^2=2^2-b^2 or

b^2=4-1/2 or b=sqrt 3.5 . The center is (h=1/2,k=1/2)

The ellipse is tilted at an angle of theta=tan^-1 (1/1)=45^0

Hence the equation of ellipse is.

{(x-h)cos theta+(y-k)sin theta}^2/a^2+{(y-k)cos theta-(x-h)sin theta}^2/b^2 =1 or

{(x-1/2)cos 45+(y-1/2)sin 45}^2/4+{(y-1/2)cos 45-(x-1/2)sin 45}^2/3.5 =1 or

{(x-1/2)/sqrt2+(y-1/2)/sqrt2}^2/4+{(y-1/2)/sqrt2-(x-1/2)/sqrt2}^2/3.5 =1

{(2x-1)/(2sqrt2)+(2y-1)/(2sqrt2)}^2/4+{(2y-1)/(2sqrt2)-(2x-1)/(2sqrt2)}^2/3.5 =1 or

{(2x-1+2y-1)/(2sqrt2)}^2/4+{(2y-2x)/(2sqrt2)}^2/3.5 =1 or

(x+y-1)^2/8+(y-x)^2/7 =1 [Ans]