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Question 4 0/1 pt 5399 Details Find an equation in standard form for the ellipse with center at (7,5), horizontal major axis of length 10 and passes through the point (8,4). Question Help: Video Submit Question Question 5 < B0/1 pt 399 Details A bridge is to be built in the shape of a semi-elliptical arch and is to have a span of 260 meters. The height of the arch at a distance of 112 meters from the center is to be 33 meters. Find the height of the arch at its center. meters Question Help: Video Submit Question

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center to the x-ans in son 4 Standard form of ellipse with h, K) and major asil foncilier (2-hid + (-x f 1 where axb al be Guren: Center (hik) = (1,5) A b = 1 and k=s. thength of major anis = 2a s lo & a=5 Theo I put all these rawes in , we get 14-5) =1 62 (2-1)2 + (5) ellipse passes through Also, it į given that point (8,4). it satisfy the equation. sor we get (8-3)? (5)2 + 14-5) =1 62 ㅗ 25 b² 24 25 62 25 b = 25 24 Thes, the ellipse is given by (2-7) + (7-5) = 25/24 (0-19%+ 24 (7-5) = 1 25 25 25

5 that the untu of ellipse at ougin 10,0) (0, b) Î (112,33) we need to find b. (-130,0) 10,0) (130,0) 260 standard form of het the ellipse center at (oo) u x2 + 72 at a 2 12 Chitren span = 24 = 260 ) from the graph, we have a = 130 Also, ellipse passes through the point (112, 33). Then I put all the in ① we we get لقبيلوح (11232 (130) + (33) b2 12544 + 16900 1089. =] b2 1089 b2 12544 16900 4356 16900 (33) x (130) (66)2 62= 1089 x 16100 4356 x (130)? (66)2 (33)2 = 38 x 130 = 65 b = t62 aich is b = 65 Thule the height of the 165 m center at its

*7 from the graph, we have Length of the transverse axis to E (-3,2) 20 = a = 5 10 center (hk) = (-3,2). -8 Then the standard (x-4) 2 - (y - k)? ba (X-4–3}}} = 6² (x+372 25 (4-232 =) since hypebola passes through 14, 1) then (4+3) (1-2) =) 25 49 25 - 봉 =) to 49,-) - 25 24 25 63 - 25, 24 (2+372 (y-2) 25 23 24 (x+3) 24 (4-2)= 25 23