Substitute y=x+ 1 into the above formula to get: (a 2+b 2) x 2+a 2+b 2) x 2+2a 2 * x+a 2-a 2 * b 2 = 0.

From kpq = 1, | pq | = 10 (1/2)/2, | x 1-x2 | = 5 (1/2)/2.

(x 1-x2)^2=(x 1+x2)^2-4x 1x2=(-2a^2)^2/(a^2+b^2)^2-4(a^2-a^2b^2)/(a^2+b^2)=5/4..................( 1)

When OP is perpendicular to OQ, we get y1/x1* y2/x2 =-1,x 1x2+y 1y2=0, that is, x1x2+(x/x2)

2x 1x2+(x 1+x2)+ 1=0,2(a^2-a^2b^2)/(a^2+b^2)-2a^2/(a^2+b^2)+ 1=0.............................(2)

The following results can be obtained from (1)(2): a 2 = 2/3, b 2 = 2 or a 2 = 2, b 2 = 2/3.

The elliptic equation is 3x 2+y 2 = 2 or x 2+3y 2 = 2.