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



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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.