Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example, 19 A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse. Given AB = 15 cm & AP = 6 cm PB = AB – AP PB = 15 – 6 PB = 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, sin2𝜃 + cos2𝜃 = 1 Putting 𝑠𝑖𝑛𝜃 = 𝑦6 and 𝑐𝑜𝑠𝜃 = 𝑥9 𝑦62 + 𝑥92 = 1 𝑦236 + 𝑥281 = 1 𝒙𝟐𝟖𝟏 + 𝒚𝟐𝟑𝟔 = 1 Hence it satisfies the equation of ellipse 𝑥2𝑎2 + 𝑦2𝑏2 = 1 Thus locus of P is ellipse Hence proved

Chapter 11 Class 11 Conic Sections

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.