Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Important

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10 Important

Misc 11 Important

Misc 12 Important

Misc 13

Misc 14 Important

Misc 15 Important

Misc 16 (MCQ)

Question 1 (a) Deleted for CBSE Board 2024 Exams

Question 1 (b) Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams You are here

Question 4 (MCQ) Important Deleted for CBSE Board 2024 Exams

Question 5 (MCQ) Important Deleted for CBSE Board 2024 Exams

Question 6 (MCQ) Deleted for CBSE Board 2024 Exams

Question 7 (MCQ) Important Deleted for CBSE Board 2024 Exams

Question 8 (MCQ) Important Deleted for CBSE Board 2024 Exams

Last updated at April 16, 2024 by Teachoo

Question 3 Show that the normal at any point ΞΈ to the curve x = a cos π + a π sin π, y = a sin π β a π cos π is at a constant distance from the origin.Given curve π=π cosβ‘π+π π sinβ‘π , π=π sinβ‘πβ π π cosβ‘π We need to show distance of a normal from (0, 0) is constant First , calculating Equation of Normal We know that Slope of tangent is ππ¦/ππ₯ π π/π π= (π π/π π½)/(π π/π π½) Finding π π/π π½ Given π₯=π cosβ‘π+π π sinβ‘π Diff. w.r.t ΞΈ ππ₯/ππ= π(π cosβ‘π + π π sinβ‘π )/ππ ππ₯/ππ= (π(π cosβ‘π))/ππ + (π(ππ sinβ‘π))/ππ ππ₯/ππ = π (βsinβ‘π )+π (π(π sinβ‘π))/ππ Using product rule (u v)β = uβ v + vβ u ππ₯/ππ=β a sin ΞΈ+a (ππ/ππ sin ΞΈ+ (π(π ππ π))/ππ ΞΈ) ππ₯/ππ=β a sin ΞΈ+a ( sin ΞΈ+ΞΈ cosβ‘γΞΈ γ ) ππ₯/ππ=β a sin ΞΈ+a sin ΞΈ+π ΞΈ cosβ‘γΞΈ γ π π/π π½=π π½ πππβ‘γπ½ γ Finding π π/π π½ Given π¦=π π ππβ‘πβπ π πππ β‘π Diff. w.r.t ΞΈ ππ¦/ππ= π(π π ππβ‘πβπ π πππ β‘π)/ππ ππ¦/ππ=a cos ΞΈβa (ππ/ππ cos ΞΈ+ (π(πππ π))/ππ ΞΈ) ππ¦/ππ=a cos ΞΈβa ( cos ΞΈβΞΈ sinβ‘γΞΈ γ ) ππ¦/ππ=a cos ΞΈβa cos ΞΈ+π ΞΈ sinβ‘γΞΈ γ π π/π π½=π π½ πππβ‘γπ½ γ Now, π π/π π= (π πβπ π½)/(π πβπ π½) ππ¦/ππ₯=(π π sinβ‘π)/(π π cosβ‘π ) ππ¦/ππ₯=sinβ‘π/cosβ‘π π π/π π= πππβ‘π½ We know that Slope of tangent of Γ Slope of normal = β1 tan ΞΈ Γ Slope of normal = β1 Slope of normal = (β1)/tanβ‘π Slope of normal =βπππβ‘π½ Equation of normal which passes through the curve π = a cos ΞΈ + a ΞΈ sin ΞΈ & π = a sin ΞΈ β a ΞΈ cos ΞΈ & has slope βπππβ‘π½ is We know that Equation of line passing through (π₯1 , π¦1) & having slope m is (π¦βπ¦1) = m(π₯βπ₯1) (π¦β(π sinβ‘πβπ cosβ‘π ))=βππ¨πβ‘π½(π₯β(π cosβ‘π+π π sinβ‘π )) (π¦βπ sinβ‘π+π π cosβ‘π )=(βπππβ‘π½)/πππβ‘π½ (π₯βπ cosβ‘π+π π sinβ‘π ) π¬π’π§β‘π½(π¦βπ sinβ‘π+π π cosβ‘π )=βππ¨π¬β‘π½(π₯βπ cosβ‘πβπ π sinβ‘π ) π¦ sinβ‘πβπ sin2 π+π π .cosβ‘π sinβ‘π=βπ₯ cosβ‘π+π cos2 π+π πsinβ‘π cosβ‘π π¦ sinβ‘πβπ sin2 π+π₯ cosβ‘πβπ cos2 π=π π sinβ‘π cosβ‘πβπ π sinβ‘π cosβ‘π π¦ sinβ‘π+π₯ cosβ‘πβπ sin2 πβπ cos2 π=0 π¦ sinβ‘π+ π₯ cosβ‘πβπ (ππππ π½+ππππ π½)=0 π¦ sinβ‘π+π₯ cosβ‘πβπ (π)=0 π πππβ‘π½+π πππβ‘π½ βπ = π We know that Distance of line ax + by + c = 0 from point (x1, y1) is d = |πππ + πππ +π|/β(π¨^π + π©^π ) Finding Distance of Normal from Origin Equation of normal is π πππβ‘π½+π πππβ‘π½ βπ = π Comparing with aπ₯ + bπ¦ + c = 0 β΄ a = cos ΞΈ, b = sin ΞΈ & c = β a We need to find distance of normal from origin i.e. ππ = 0 & ππ = 0 π= |cosβ‘γπ(0) + sinβ‘γπ(0) β πγ γ |/β(cos^2β‘π + sin^2β‘π ) d = |0 + 0 β π|/β1 d = |βπ|/1 d = π/1 d = a d = Constant Hence, distance of normal from origin is a constant. Hence proved.