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Misc 2 - An arch is in form of a parabola, its axis vertical - Parabola - Arch/mirror problem

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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise
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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 ﷐﷐𝟓﷮𝟐﷯, −𝟏𝟎﷯ 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 = ﷐𝟓﷮𝟑𝟐﷯ Hence, the equation of parabola is x2 = –4ay Putting a = ﷐5﷮32﷯ x2 = –4﷐﷐5﷮32﷯﷯y x2 = – ﷐𝟓𝒚﷮𝟖﷯ 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 = ﷐−5𝑦﷮8﷯ Putting y = –2 x2 = –4 × ﷐5﷮32﷯ × (–2) x2 = ﷐4 × 5 × 2﷮32﷯ x2 = ﷐5﷮4﷯ x = ± ﷐﷮﷐5﷮4﷯﷯ x = ± ﷐﷐﷮𝟓﷯﷮𝟐﷯ Hence, PN = ﷐﷐﷮5﷯﷮2﷯ So, MN = 2PN = 2 ×﷐﷐﷮5﷯﷮2﷯ = ﷐﷮5﷯ Thus, the width of the arch MN = ﷐﷮𝟓﷯ m = 2.23 m (approx.)

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