The shape of the Gateway arch in st Louis Missoutl can be ap

The shape of the Gateway arch in st, Louis, Missoutl, can be approximately modeled with the equations of two parabolas: one parabola for the outer upper surface. and one for the inner/lower surface. The geight of the arch narrows as it rises. Therefore, the insides of the bases are only 540 feet at ground level, but the inside of the arch has a beight of 615 feet, Suppose you are standing at teh origin which is at ground level directly undereath the center of the arch. a. Find the verted,and x-intercepts for both the outer and inner parabolas. Outer parabola: vertex: x-intercepts: Inner parabola: verted: x-intercepts: b. Find the beitht above the ground of the focus and directrix for each parabola. Round your answers to the nearest tenth of a foot. Outer parabola: focus height: directrix height: Inner parabola: focus height: directrix height: c. Find the equation for both the outer and inner parabolas in standard form.

Solution

We are standing at the origin which is at the ground level directly beneath the center of the arch. Then, the ground is the X-Axis and Y -Axis is the line of symmetry . Let the equations of the parabolas representing the outer and the inner arches be y = a(x-h)² + k ...(1) and y =b( x-p)² + q...(2) respectively. Since the y-Axis is the line of symmetry for the two parabolas, the x-coordinates of the vertices are 0 so that h = 0 and p = 0. Also, the y -coordinates of the vertices of the two parabolas are 630 and 615 respective. Thus, the vertices of the two parabolas are (0, 630) and (0, 615) respectively. Then the equations of the outer and the inner parabolas change to y = ax² + 630...(3) and y = bx² + 615...(4) respectively. Further, since the outside and the inside bases are 630 and 540 feet apart respectively, the outer parabola passes through the points ( - 315, 0) and ( 315,0) and the inner parabola passes through the points ( -270,0) and (270, 0) . On substituting x = 315 and y = 0 in the 3rd equation, we get a(315)² + 630 = 0 or, 99225a = - 630 so that a = -630/99225 = -0.00635 (approximately). Similarly, on substituting x = 270 and y = 0 in the 4th equation, we get b(270)²+615 = 0 or, 72900b = -615 so that b = -615/72900 = - 0.00844 (approx.). Thus, the equations of the outer and the inner parabolas are y = - 0.00635x² + 630...(5) and y = - 0 00844x² + 615....(6) respectively. We know that the x-intercept is where y = 0. Therefore, the x intercepts for the outer parabola are -315 and 315. Its vertex is at (0, 630). For, the inner parabola, the x-intercepts are -270 and 270. Its vertex is at (0, 615).   

b. We know for the parabola (x –h)² = 4p( y –k), the focus is (h, k + p) and the directrix is the line y = k-p. For the outer parabola, the equation is (x-0)² = (y - 630)/ ( -0.00635). Therefore, 4p = -1/0.00635 and hence p = -1/(4*0.00635) = -1/0.0254 = -39.370 .Thus, the focus of the outer parabola is at (0,590.63 ) and its directrix is the line y = 630-( -39.370) or, y = 669.37. Similarly, since the equation of the inner parabola is (x – 0)² = ( y- 615)/ (- 0.00844). Therefore, here 4p = -1/0.00844 so that p = - 1/( 4*0.00844) = -1/0.03376 = -29.62. Therefore, the focus of the inner parabola is at (0, 585.38 ). Its directrix is the line y = 615- (-29.62) or, y = 644.62. Thus, for the outer parabola, the focus height is 590.63feet. Its directrix height is 669.37 feet. For the inner parabola, the focus height is 585.38 feet. Its directrix height is 644.62 feet.

c. The vertex form of the equation of the outer parabola is y = - 0.00635x² + 630. This also its equation in the standard form. Similarly, the vertex form of the equation of the inner parabola is y= - 0 00844x² + 615. This also its equation in the standard form