Question

Given the ellipse, [(x + 5)² ÷ 25)] + [(y - 3)² ÷ 49] = 1, determine the location of the centre, vertices, and foci, and then sketch the graph based on those attributes.

Answer 1

* The of the given ellipse is equation (x+5) 2 (y-3 1 25 49 this equation with Comparing (-u) ² 2 (y-k² we =1 get + az are The centre of the ellipse is (-5, 3) = (hak) The co-ordinates of the vertices Chakta) = (-5,3 +7) [as a=7 (-5,10) and (-5,-4) k= 3 So Two vertices (-5, 1o) and (-5,-4). The co-ordinates of the foci are (h. k I c), where h =-5 3 are e ² = a² - 6 = 49-25=24 2. = 24 256. Then, (h, k+ c) = (-5, 3 + 256) foci are (-5,3 +256 3 +256) and so, Two (-5, 3-256)