Chapter 10 Class 11 Conic Sections
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Misc 2 - An arch is in form of a parabola, its axis vertical - Parabola - Arch/mirror problem

Misc 2 - Chapter 11 Class 11 Conic Sections - Part 2
Misc 2 - Chapter 11 Class 11 Conic Sections - Part 3 Misc 2 - Chapter 11 Class 11 Conic Sections - Part 4 Misc 2 - Chapter 11 Class 11 Conic Sections - Part 5

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Transcript

Misc 2 An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? Arch is downwards, Since, the axis of parabola is negative y-axis, its equation is x2 = 4ay First, we find coordinates of point B Given, Width of Parabola = AB = 5m So, BC = 5 2 m Also, Parabola is 10 m high Hence, OC = 10 m BD = OC = 10m Hence, coordinates of point B is , Now since point B 5 2 , 10 lies on the parabola It will satisfy the equation of the parabola, Putting x = 5 2 , y = 10 in equation x2 = 4ay 5 2 2 = 4a ( 10) 25 4 1 4 1 10 = a 25 4 1 4 1 10 = a 5 4 4 2 = a 5 32 = a a = Hence, the equation of parabola is x2 = 4ay Putting a = 5 32 x2 = 4 5 32 y x2 = We need to find how wide the arch is it 2 m from the vertex of the parabola Let point P be point 2 m from vertex of parabola Hence, OP = 2m We need to find MN Now equation of parabola is x2 = 5 8 Putting y = 2 x2 = 4 5 32 ( 2) x2 = 4 5 2 32 x2 = 5 4 x = 5 4 x = Hence, PN = 5 2 So, MN = 2PN = 2 5 2 = 5 Thus, the width of the arch MN = m = 2.23 m (approx.)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.