7) Find the center , the verticies, and the foci of the ellipse. The.pdf

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7) Find the center , the verticies, and the foci of the ellipse. Then draw the graph. 25x^2 +16y^2-50x+64y-311=0 8) Use verticies and asymptotes to graph the hyperbola 9x^2-4y^2 = 36 Solution we can write the above equation as (x-1)2 / 16 + (y+2) 2 /25 = 1 the origin is shifted to (1,-2) vertices are (1,-2) now as the vertical radius is larger than horizontal radius the x coordinate of focus would be - 2 and the y coordinate would be +-3 from the y cordinate so the focii are (1,-5) (1,1)) when we draw the graph everything remains the same but the vertices are shifted b)9x^2-4y^2 = 36 , x^2/4-y^2/9 = 1 a = 2 , b = 3 x = +-(2/3)y are the equations of asymptotes.

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