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4. [-/0.12 Points] DETAILS SCALCET8 10.5.010. 0/100 Submissions Used Consider the following. 2 1.5 y 1- 0.5- 3 -0.5 Find an equation of the parabola. Find the focus and directrix. focus (x, y) = directrix3. [-/0.12 Points] DETAILS SCALCET8 10.5.004. 0/100 Submissions Used Find the vertex, focus, and directrix of the parabola. 3x2 + 12y = 0 vertex ( x , y ) = focus ( x , y ) = directrix Sketch its graph. N 4 2 -4 -2 2 I X - 2 1 2 3 - 2 -2 -3 -4 -4 O O w 2 N - X -4 -3 -2 -1 2 -4 -2 2 -2 -21 O O1. [-/0.12 Points] DETAILS SCALCET8 10.5.AE.003. 0/100 Submissions Used MY NOTES ASK YOUR TEACHER Video Example () EXAMPLE 3 Find the equation of the ellipse with foci (0, +2) and vertices (0, +3). SOLUTION Using the standard notation, we have c = and a = . Then we obtain b2 = a2 - c2 = 9 - so an equation of the ellipse is Another way of writing this equation is 9x2 + 5y2 = Need Help? Read It 2. [-/0.12 Points] DETAILS SCALCET8 10.5.AE.007. 0/100 Submissions Used MY NOTES ASK YOUR TEACHER EXAMPLE 7 Sketch the conic below and find its foci. 36x2 - 25y2 - 288x + 100y + 1376 = 0 SOLUTION We complete the squares as follows. 25(12 - 4y) - 36(x2 - 8x) = 1376 25 ( 1 2 - 4y + - 36 (x2 - 8x + = 1376 + 100 - 576 25 (y -)2 - 36(x - 1 2 ( x - )2 - ( x -1 Video Example () This is in the standard form except that x and y are replaced by x - 4 and y - 2. Thus a2 = , 62 = , and c2 = . The hyperbola is shifted four units to the right and two units upward. The foci are (x, y) = ( and (4, 2 - V61) and the vertices are (x, y) = and (4 , -4 ). The asymptotes are y - 2 = + (x * (x - 4). The hyperbola is sketched in the figure. Need Help? Read It