Question 2 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

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

(A) touch each other

(B) cut at right angle

(C) cut at an angle π/3

(D) cut at an angle π/4

Transcript

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/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
𝑥^3−3𝑥𝑦^2+2=0
Differentiating w.r.t x
〖3𝑥〗^2− 3𝑦^2 − 6xy 𝑑𝑦/𝑑𝑥 = 0
〖3𝑥〗^2− 3𝑦^2 = 6xy 𝑑𝑦/𝑑𝑥
𝑑𝑦/𝑑𝑥 = (3𝑥^2 −3𝑦^2)/6𝑥𝑦
𝒅𝒚/𝒅𝒙 = (𝒙^𝟐 −𝒚^𝟐)/𝟐𝒙𝒚
Let m1 = (𝑥^2 −𝑦^2)/2𝑥𝑦
Finding slope of tangent of second curve
3𝑥^2 𝑦−𝑦^3−2=8
Differentiating w.r.t x
3𝑥^2 𝑑𝑦/𝑑𝑥+6𝑥𝑦 − 3𝑦^2 𝑑𝑦/𝑑𝑥 = 0
(3𝑥^2−3𝑦^2 ) 𝑑𝑦/𝑑𝑥 = −6xy
𝑑𝑦/𝑑𝑥 = (−6𝑥𝑦)/(3𝑥^2 −3𝑦^2 )
𝒅𝒚/𝒅𝒙 = (−𝟐𝒙𝒚)/(𝒙^𝟐 −𝒚^𝟐 )
Let, 𝒎_𝟐= (−2𝑥𝑦)/(𝑥^2 −𝑦^2 )
Finding Product of m1 & m2
m1 × m2 = (𝑥^2 − 𝑦^2)/2𝑥𝑦 × ((−2𝑥𝑦)/(𝑥^(2 )− 𝑦^2 ))
= −1
Since, product of the slopes is −1
∴ Angle between tangents is 90°
Thus, curves cut each other at right angle.
So, the correct answer is (B)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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