# SOLUTION: How do I find the vertex, focus, directrix, axis of symmetry, and latus rectum of this parabola equation? y - 1 = ¼(x + 2)²

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 421654: How do I find the vertex, focus, directrix, axis of symmetry, and latus rectum of this parabola equation? y - 1 = ¼(x + 2)² Answer by Edwin McCravy(8879)   (Show Source): You can put this solution on YOUR website!How do I find the vertex, focus, directrix, axis of symmetry, and latus rectum of this parabola equation? y - 1 = ¼(x + 2)² ``` y - 1 = ¼(x + 2)² Multiply both sides by 4 4*(y - 1) = 4*¼(x + 2)² 4*(y - 1) = 1(x + 2)² 4(y - 1) = (x + 2)² Swap left and right sides: (x + 2)² = 4(y - 1) Compare to this standard form: (x - h)² = 4p(y - k) which has these properties: 1. vertex is the point (h,k) 2. line of symmetry equation is x = h 3. focus is the point (h,k+p) 4. directrix is the line whose equation is y = k-p 5. length of latus rectum = |4p| h = -2, k = 1, 4p = 4, so p = 1 So for this parabola, 1. vertex is the point (h,k) = (-2,1) 2. line of symmetry equation is x = h or x = -2 3. focus is the point (h,k+p) = (-2,1+1) = (-2,2) 4. directrix is the line whose equation is y = k-p or y = 1-1 or y = 0, which is the x-axis. 5. length of latus rectum = |4p| = 4 We plot the vertex (-2,1), focus (-2,2), the line of symmetry (in green), and the directrix y = 0 happens to be the the x-axis: We draw the latus rectum which is a horizontal line segment 4p or 4 units long going through and centered on the focus: Finally we draw the parabola through the ends of the latus rectum and through the vertex: Edwin```