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Misc, 5 A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis. Let AB be the rod where A touches the x-axis & B touches the y-axis Let point P(x, y) Given AB = Length of rod = 12 cm & AP = 3 cm PB = AB – AP PB = 12 – 3 = 9cm Drawing PQ ⊥ BO and PR ⊥ OA Hence, PQ = x & PR = y Let ∠ PAR = θ Now, PQ & AO are parallel lines (As both are perpendicular to y-axis) & BA is the transversal So ∠ BPQ = ∠ PAR = θ Now we know that, ﷐ sin2﷮𝜃﷯ + ﷐cos2﷮𝜃﷯ = 1 Putting ﷐ 𝑠𝑖𝑛﷮𝜃﷯ = ﷐𝑦﷮6﷯ and ﷐𝑐𝑜𝑠﷮𝜃﷯ = ﷐𝑥﷮9﷯ ﷐﷐﷐𝑦﷮3﷯﷯﷮2﷯ + ﷐﷐﷐𝑥﷮9﷯﷯﷮2﷯ = 1 ﷐𝑦2﷮9﷯ + ﷐𝑥2﷮81﷯ = 1 ﷐𝒙𝟐﷮𝟖𝟏﷯ + ﷐𝒚𝟐﷮𝟗﷯ = 1 Hence it satisfies the equation of ellipse ﷐﷐𝑥﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐𝑦﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 Thus locus of P is ellipse

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo