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  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise

Transcript

Misc 23 Prove that the product of the lengths of the perpendiculars drawn from the points (√(π‘Ž^2 βˆ’ 𝑏^2 ), 0) and ( βˆ’ √(π‘Ž^2 βˆ’ 𝑏^2 ), 0) to the line π‘₯/π‘Ž cos πœƒ + 𝑦/𝑏 sin πœƒ = 1 is b2 . Let point A be (√(π‘Ž^2βˆ’π‘^2 ), 0) & point B be (βˆ’βˆš(π‘Ž^2βˆ’π‘^2 ), 0) The given line is π‘₯/π‘Ž cos ΞΈ + 𝑦/𝑏 sinΞΈ = 1 We need to show that Product of the length of perpendiculars from point A & point B to the line π‘₯/π‘Ž cos ΞΈ + 𝑦/𝑏 sin ΞΈ = 1 is b2 Let p1 be the perpendicular distance from point A(√(π‘Ž^2 βˆ’ 𝑏^2 ), 0) to the line π‘₯/π‘Ž cos ΞΈ + 𝑦/π‘Ž sin ΞΈ = 1 & p2 be the perpendicular distance from point B( βˆ’ √(π‘Ž^2 βˆ’ 𝑏^2 ), 0) to the line π‘₯/π‘Ž cos ΞΈ + 𝑦/π‘Ž sin ΞΈ = 1 We need to show p1 Γ— p2 = b2 Calculating p1 & p2 Given line is (π‘₯ )/π‘Ž cos ΞΈ + 𝑦/𝑏sin ΞΈ = 1 (cosβ‘πœƒ/π‘Ž) x + (sinβ‘πœƒ/𝑏)y βˆ’ 1 = 0 The above equation is of the form Ax + By + C = 0 Here A = cosβ‘πœƒ/π‘Ž, B = sinβ‘πœƒ/𝑏 & C = βˆ’ 1 We know that Distance of a point (x1, y1) from line Ax + By + C = 0 is d = |〖𝐴π‘₯γ€—_1 + 〖𝐡𝑦〗_1 + 𝑐|/√(𝐴^2 + 𝐡^2 ) Perpendicular distance p1 of point A(√(π‘Ž^2 βˆ’ 𝑏^2 ), 0) from the line π‘₯/π‘Ž cos ΞΈ + 𝑦/𝑏 sin ΞΈ = 1 is d = |〖𝐴π‘₯γ€—_1 + 〖𝐡𝑦〗_1 + 𝑐|/√(𝐴^2 + 𝐡^2 ) Putting values p1 = |cosβ‘πœƒ/π‘Ž √(π‘Ž^2 + 𝑏^2 ) + sinβ‘πœƒ/𝑏 (0) + ( βˆ’ 1)|/√((cosβ‘πœƒ/π‘Ž)^2 + (sinβ‘πœƒ/𝑏)^2 ) p1 = (√(π‘Ž^2 βˆ’ 𝑏^2 )/π‘Ž cosβ‘πœƒ + 0 βˆ’ 1)/√((γ€–π‘π‘œπ‘ γ€—^2 πœƒ)/π‘Ž^2 + (〖𝑠𝑖𝑛〗^2 πœƒ)/𝑏^2 ) p1 = (√(π‘Ž^2 βˆ’ 𝑏^2 )/π‘Ž cosβ‘πœƒ βˆ’ 1)/√((γ€–π‘π‘œπ‘ γ€—^2 πœƒ)/π‘Ž^2 + (〖𝑠𝑖𝑛〗^2 πœƒ)/𝑏^2 ) Perpendicular distance p2 of point B = ( βˆ’ √(π‘Ž^2 βˆ’ 𝑏^2 ), 0) from the line π‘₯/π‘Ž cos ΞΈ + 𝑦/𝑏 sin ΞΈ = 1 is d = |𝐴π‘₯1 + 𝐡𝑦1 + 𝑐|/√(π‘Ž^2 + 𝑏^2 ) Putting values p2 = |cosβ‘πœƒ/π‘Ž ( βˆ’ √(π‘Ž^2 βˆ’ 𝑏^2 )) + sinβ‘πœƒ/𝑏 (0) + ( βˆ’ 1)|/√((cosβ‘πœƒ/π‘Ž)^2 + (sinβ‘πœƒ/𝑏)^2 ) p2 = ( βˆ’ √(π‘Ž^2 βˆ’ 𝑏^2 )/π‘Ž cosβ‘πœƒ + 0 βˆ’ 1)/√((cosβ‘πœƒ/π‘Ž)^2 + (sinβ‘πœƒ/𝑏)^2 ) p2 = ( βˆ’ √(π‘Ž^2 βˆ’ 𝑏^2 )/π‘Ž cosβ‘πœƒ βˆ’ 1)/√(γ€–π‘π‘œπ‘ γ€—^2/π‘Ž^2 + (〖𝑠𝑖𝑛〗^2 πœƒ)/𝑏^2 ) Finding p1p2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.