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Last updated at March 30, 2023 by Teachoo

Misc 23 Prove that the product of the lengths of the perpendiculars drawn from the points (β(π^2 β π^2 ), 0) and ( β β(π^2 β π^2 ), 0) to the line π₯/π cos π + π¦/π sin π = 1 is b2 . Let point A be (β(π^2βπ^2 ), 0) & point B be (ββ(π^2βπ^2 ), 0) The given line is π₯/π cos ΞΈ + π¦/π sinΞΈ = 1 We need to show that Product of the length of perpendiculars from point A & point B to the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 is b2 Let p1 be the perpendicular distance from point A(β(π^2 β π^2 ), 0) to the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 & p2 be the perpendicular distance from point B( β β(π^2 β π^2 ), 0) to the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 We need to show p1 Γ p2 = b2 Calculating p1 & p2 Given line is (π₯ )/π cos ΞΈ + π¦/πsin ΞΈ = 1 (cosβ‘π/π) x + (sinβ‘π/π)y β 1 = 0 The above equation is of the form Ax + By + C = 0 Here A = cosβ‘π/π, B = sinβ‘π/π & C = β 1 We know that Distance of a point (x1, y1) from line Ax + By + C = 0 is d = |γπ΄π₯γ_1 + γπ΅π¦γ_1 + π|/β(π΄^2 + π΅^2 ) Perpendicular distance p1 of point A(β(π^2 β π^2 ), 0) from the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 is d = |γπ΄π₯γ_1 + γπ΅π¦γ_1 + π|/β(π΄^2 + π΅^2 ) Putting values p1 = |cosβ‘π/π β(π^2 + π^2 ) + sinβ‘π/π (0) + ( β 1)|/β((cosβ‘π/π)^2 + (sinβ‘π/π)^2 ) p1 = (β(π^2 β π^2 )/π cosβ‘π + 0 β 1)/β((γπππ γ^2 π)/π^2 + (γπ ππγ^2 π)/π^2 ) p1 = (β(π^2 β π^2 )/π cosβ‘π β 1)/β((γπππ γ^2 π)/π^2 + (γπ ππγ^2 π)/π^2 ) Perpendicular distance p2 of point B = ( β β(π^2 β π^2 ), 0) from the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 is d = |π΄π₯1 + π΅π¦1 + π|/β(π^2 + π^2 ) Putting values p2 = |cosβ‘π/π ( β β(π^2 β π^2 )) + sinβ‘π/π (0) + ( β 1)|/β((cosβ‘π/π)^2 + (sinβ‘π/π)^2 ) p2 = ( β β(π^2 β π^2 )/π cosβ‘π + 0 β 1)/β((cosβ‘π/π)^2 + (sinβ‘π/π)^2 ) p2 = ( β β(π^2 β π^2 )/π cosβ‘π β 1)/β(γπππ γ^2/π^2 + (γπ ππγ^2 π)/π^2 ) Finding p1p2