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Misc 23 - Prove that product of lengths of perpendiculars

Misc 23 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 23 - Chapter 10 Class 11 Straight Lines - Part 3
Misc 23 - Chapter 10 Class 11 Straight Lines - Part 4
Misc 23 - Chapter 10 Class 11 Straight Lines - Part 5
Misc 23 - Chapter 10 Class 11 Straight Lines - Part 6
Misc 23 - Chapter 10 Class 11 Straight Lines - Part 7
Misc 23 - Chapter 10 Class 11 Straight Lines - Part 8

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.