NCERT Exemplar - MCQs

Chapter 6 Class 12 Application of Derivatives
Serial order wise

## (D) cut at an angle Ο/4

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### Transcript

Question 2 The two curves x3 β 3xy2 + 2 = 0 and 3x2y β y3 = 2 (A) touch each other (B) cut at right angle (C) cut at an angle π/3 (D) cut at an angle π/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₯^3β3π₯π¦^2+2=0 Differentiating w.r.t x γ3π₯γ^2β 3π¦^2 β 6xy ππ¦/ππ₯ = 0 γ3π₯γ^2β 3π¦^2 = 6xy ππ¦/ππ₯ ππ¦/ππ₯ = (3π₯^2 β3π¦^2)/6π₯π¦ ππ/ππ = (π^π βπ^π)/πππ Let m1 = (π₯^2 βπ¦^2)/2π₯π¦ Finding slope of tangent of second curve 3π₯^2 π¦βπ¦^3β2=8 Differentiating w.r.t x 3π₯^2 ππ¦/ππ₯+6π₯π¦ β 3π¦^2 ππ¦/ππ₯ = 0 (3π₯^2β3π¦^2 ) ππ¦/ππ₯ = β6xy ππ¦/ππ₯ = (β6π₯π¦)/(3π₯^2 β3π¦^2 ) ππ/ππ = (βπππ)/(π^π βπ^π ) Let, π_π= (β2π₯π¦)/(π₯^2 βπ¦^2 ) Finding Product of m1 & m2 m1 Γ m2 = (π₯^2 β π¦^2)/2π₯π¦ Γ ((β2π₯π¦)/(π₯^(2 )β π¦^2 )) = β1 Since, product of the slopes is β1 β΄ Angle between tangents is 90Β° Thus, curves cut each other at right angle. So, the correct answer is (B)