Misc 8 - An equilateral triangle is inscribed in parabola - Parabola - Triangle in parabola problem

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Misc 8 An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle. Let length of equilateral triangle be s Hence OA = OB = AB = s Here, OC ⊥ AB So, ∠ OCA = ∠ OCB = 90° And AC = BC So, AC = BC = ﷐𝐴𝐵﷮2﷯ AC = BC = ﷐𝑠﷮2﷯ We find coordinates of point B Now, in right triangle ∆OBC Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 (OB)2 = (OC)2 + (BC)2 s2 = (OC)2 + ﷐﷐﷐𝑠﷮2﷯﷯﷮2﷯ s2 = (OC)2 + ﷐﷐𝑠﷮2﷯﷮4﷯ s2 – ﷐﷐𝑠﷮2﷯﷮4﷯ = (OC)2 ﷐4﷐𝑠﷮2﷯ − ﷐𝑠﷮2﷯﷮4﷯ = (OC)2 ﷐3﷐𝑠﷮2﷯﷮4﷯ = (OC)2 (OC)2 = ﷐3﷐𝑠﷮2﷯﷮4﷯ OC = ﷐﷮﷐3﷐𝑠﷮2﷯﷮4﷯﷯ OC = ﷐﷐﷮3﷯𝑠﷮2﷯ Hence coordinate of point B is B(﷐﷐﷮𝟑﷯𝒔﷮𝟐﷯,﷐𝒔﷮𝟐﷯) Now, point B lies on parabola So, it must satisfy its equation Putting x = ﷐﷐﷮3﷯𝑠﷮2﷯, y = ﷐𝑠﷮2﷯ in equation of parabola y2 = 4ax ﷐﷐﷐𝑠﷮2﷯﷯﷮2﷯ = 4a(﷐﷐﷮3﷯𝑠﷮2﷯) ﷐﷐𝑠﷮2﷯﷮4﷯ = 4a(﷐﷐﷮3﷯𝑠﷮2﷯) ﷐﷐𝑠﷮2﷯﷮𝑠﷯ = 4 × 4a(﷐﷐﷮3﷯﷮2﷯) s = 8﷐﷮𝟑﷯ a Hence, side of equilateral triangle = 8﷐﷮3﷯ a

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