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Last updated at March 30, 2020 by Teachoo
Transcript
Ex 10.3, 16 If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2 The given lines are x cos θ – y sin θ = k cos 2θ x sec θ + y cosec θ = k cos 2θ The distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Taking (1) x cos θ – y sin θ = k cos 2θ x cos θ – y sin θ – k cos 2θ = 0 The above equation is of the form Ax + By + C = 0 Where A = cos θ, B = – sin θ, C = – k cos 2θ Also, p is the perpendicular distance from origin to line (1) Here x1 = 0, y1 = 0 , & d = p Putting values p = |cos〖θ(0) + (−sinθ )(0) − k cos2θ 〗 |/√((cosθ )^2 + (−sinθ )2) Taking (1) x cos θ – y sin θ = k cos 2θ x cos θ – y sin θ – k cos 2θ = 0 The above equation is of the form Ax + By + C = 0 Where A = cos θ, B = – sin θ, C = – k cos 2θ Also, p is the perpendicular distance from origin to line (1) Here x1 = 0, y1 = 0 , & d = p Putting values p = |cos〖θ(0) + (−sinθ )(0) − k cos2θ 〗 |/√((cosθ )^2 + (−sinθ )2) Taking (1) x cos θ – y sin θ = k cos 2θ x cos θ – y sin θ – k cos 2θ = 0 The above equation is of the form Ax + By + C = 0 Where A = cos θ, B = – sin θ, C = – k cos 2θ Also, p is the perpendicular distance from origin to line (1) Here x1 = 0, y1 = 0 , & d = p Putting values p = |cos〖θ(0) + (−sinθ )(0) − k cos2θ 〗 |/√((cosθ )^2 + (−sinθ )2) x sin θ + y cos θ = k cos θ sin θ sin θ x + cos θ y – k cos θ sin θ = 0 The above equation is of the form Ax + By + C = 0 where A = sin θ , B = cos θ , C = –k cos θ sin θ The distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Given that q is the perpendicular distance from origin to line (2) Here x1 = 0 , y1 = 0 , & d = q Putting values q = |sin〖θ(0) + 〖 cos〗〖θ(0) − k sin〖θ cosθ 〗 〗 〗 |/√(sin^2 θ + cos^2 θ ) q = |0 + 0 − 𝑘 sin〖𝜃 cos𝜃 〗 |/√1 q = |− 𝑘 sin〖𝜃 cos𝜃 〗 |/1 q = k sin θ cos θ We need to show p2 + 4q2 = k2 Taking LHS p2 + 4q2 Putting values from (3) & (4) = (k cos 2θ)2 + (2k sin θ cos θ)2 = k2 (cos 2θ)2 + k2(2 sin θ cos θ)2 = k2 cos2 2θ + k2 (sin 2θ)2 = k2 cos2 2θ + k2 sin2 2θ = k2 (cos2 2θ + sin2 2θ) = k2 (1) = k2 = R.H.S Hence L.H.S = R.H.S Hence proved As sin2 θ + cos2 θ = 1 Replacing θ with 2θ sin2 2θ + cos2 2θ = 1
Ex 10.3
Ex 10.3, 1 (ii) Important
Ex 10.3, 1 (iii)
Ex 10.3, 2 (i)
Ex 10.3, 2 (ii)
Ex 10.3, 2 (iii) Important
Ex 10.3, 3 (i)
Ex 10.3, 3 (ii)
Ex 10.3, 3 (iii) Important
Ex 10.3, 4
Ex 10.3, 5 Important
Ex 10.3, 6 (i) Important
Ex 10.3, 6 (ii)
Ex 10.3, 7
Ex 10.3, 8 Important
Ex 10.3, 9 Important
Ex 10.3, 10
Ex 10.3, 11
Ex 10.3, 12 Important
Ex 10.3, 13
Ex 10.3, 14 Important
Ex 10.3, 15
Ex 10.3, 16 Important You are here
Ex 10.3, 17 Important
Ex 10.3, 18 Important
Ex 10.3
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