Misc 23 - Prove that product of lengths of perpendiculars - Distance of a point from a line

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Misc 23 Prove that the product of the lengths of the perpendiculars drawn from the points (root a^2-b^2,0) and (-root a^2-b^2,0) to the line x/a cos theta +y/b sin theta =1 = b2 Let p1 be the perpendicular distance from point A(√(𝑎^2 − 𝑏^2 ), 0) to the line 𝑥/𝑎cos θ + 𝑦/𝑎 sin θ = 1 & p2 be the perpendicular distance from point B( − √(𝑎^2 − 𝑏^2 ), 0) to the line 𝑥/𝑎cos θ + 𝑦/𝑎sin θ = 1 We need to show p1 × p2 = b2 Calculating p1 & p2 Given line is 𝑥/𝑎cos θ + 𝑦/𝑏sin θ = 1 (cos⁡𝜃/𝑎)x + (sin⁡𝜃/𝑏)y − 1 = 0 = (𝑎^2 〖𝑠𝑖𝑛〗^2 𝜃 + 𝑏^2 〖𝑐𝑜𝑠〗^2 𝜃)/(𝑏^2 〖𝑐𝑜𝑠〗^2 𝜃 + 𝑎^2 〖𝑠𝑖𝑛〗^2 𝜃) × 𝑏^2 = (𝑏^2 〖𝑐𝑜𝑠〗^2 𝜃 + 𝑎^2 〖𝑠𝑖𝑛〗^2 𝜃)/(𝑏^2 〖𝑐𝑜𝑠〗^2 𝜃 + 𝑎^2 〖𝑠𝑖𝑛〗^2 𝜃) × 𝑏^2 = 𝑏^2 Hence proved

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