Check sibling questions

The two curves x 3 – 3xy 2 + 2 = 0 and 3x 2 y – y 3 = 2

(A) touch each other Β 

(B) cut at right angle

(C) cut at an angle Ο€/3Β 

(D) cut at an angle Ο€/4




Question 2 The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 = 2 (A) touch each other (B) cut at right angle (C) cut at an angle πœ‹/3 (D) cut at an angle πœ‹/4 Angles between two curves is same as angle between their tangents. So, first we will find slope of their tangents. Finding slope of tangent of first curve π‘₯^3βˆ’3π‘₯𝑦^2+2=0 Differentiating w.r.t x γ€–3π‘₯γ€—^2βˆ’ 3𝑦^2 βˆ’ 6xy 𝑑𝑦/𝑑π‘₯ = 0 γ€–3π‘₯γ€—^2βˆ’ 3𝑦^2 = 6xy 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (3π‘₯^2 βˆ’3𝑦^2)/6π‘₯𝑦 π’…π’š/𝒅𝒙 = (𝒙^𝟐 βˆ’π’š^𝟐)/πŸπ’™π’š Let m1 = (π‘₯^2 βˆ’π‘¦^2)/2π‘₯𝑦 Finding slope of tangent of second curve 3π‘₯^2 π‘¦βˆ’π‘¦^3βˆ’2=8 Differentiating w.r.t x 3π‘₯^2 𝑑𝑦/𝑑π‘₯+6π‘₯𝑦 βˆ’ 3𝑦^2 𝑑𝑦/𝑑π‘₯ = 0 (3π‘₯^2βˆ’3𝑦^2 ) 𝑑𝑦/𝑑π‘₯ = βˆ’6xy 𝑑𝑦/𝑑π‘₯ = (βˆ’6π‘₯𝑦)/(3π‘₯^2 βˆ’3𝑦^2 ) π’…π’š/𝒅𝒙 = (βˆ’πŸπ’™π’š)/(𝒙^𝟐 βˆ’π’š^𝟐 ) Let, π’Ž_𝟐= (βˆ’2π‘₯𝑦)/(π‘₯^2 βˆ’π‘¦^2 ) Finding Product of m1 & m2 m1 Γ— m2 = (π‘₯^2 βˆ’ 𝑦^2)/2π‘₯𝑦 Γ— ((βˆ’2π‘₯𝑦)/(π‘₯^(2 )βˆ’ 𝑦^2 )) = βˆ’1 Since, product of the slopes is βˆ’1 ∴ Angle between tangents is 90Β° Thus, curves cut each other at right angle. So, the correct answer is (B)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.