So AC = (? 1，y0？ b2a)，AD=(？ 1,? y0？ b2a)。

Because ∠ CAD = 90, AC? AD=0。

So 1=y02 = Y02? B4a2, that is, b2(a2? 1)a2？ b4a2= 1，

Simultaneous a2=b2+4 gives a2=6, so B2 = 2.

The elliptic equation is x, praised and trampled. What is your evaluation of this answer? Put away comments//high quality or satisfaction or special types or recommended answers dottimewindow.iperformance & window.iperformance.mark ('c _ best',+newdate); Lawyer recommendation service: If your problem has not been solved, please describe your problem in detail and consult other similar problems free of charge through Baidu Law Pro 2022- 1 1-25. It is known that the equation of ellipse C is x2/a2+y2/B2 = 1 (a > b > 0), and the left and right focal points are respectively f 65438+. The sum of the distances from the point on b>0 to its two focal points is 4, and the passing point is (0,1) 2012-01-31. As shown in the figure, ellipse C: x2/A2+Y2/B2 = 1, (a >: b >; 0) is F 1, F2, and its upper vertex is132013-11. It is known that the equation of ellipse C is X2/a2+Y2/b2= 1, and the left and right focal points are m, which is a point on the ellipse that satisfies the angle of 62014-01-kloc-0/7. It is known that the equation of ellipse C is x2/a2+y2/B2 = 1 (a > b > 0), the left and right focal points are F 1 and F2 respectively, and the focal length is 4. At point 720 17-08-09, it is known that the focal length of ellipse c: x2/A2+Y2/B2 = 1 (a > b > 0) is 2, the coordinate of the right focal point f of ellipse c is (√ 3,0), and the short axis length is 2.220/kloc-. Ellipse x2/8+y2/B2 = 1 (0 < b