# Ex 6.3,19 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.3,19 Find the points on the curve π₯^2+π¦^2 β2π₯ β3=0 at which the tangents are parallel to the π₯βππ₯ππ Given tangent is parallel to the π₯βππ₯ππ β Slope of tangent = Slope of π₯βππ₯ππ We know that Slope of tangent is ππ¦/ππ₯ π₯^2+π¦^2 β2π₯ β3=0 Differentiating w.r.t.π₯ π(π₯^2 + π¦^2 β2π₯ β3)/ππ₯=0 π(π₯^2 )/ππ₯+π(π¦^2 )/ππ₯βπ(2π₯)/ππ₯βπ(3)/ππ₯=0 2π₯+π(π¦^2 )/ππ₯ Γ ππ¦/ππ¦β2β0=0 2π₯+π(π¦^2 )/ππ₯ Γ ππ¦/ππ¦β2=0 π(π¦^2 )/ππ₯ Γ ππ¦/ππ₯=2β2π₯ 2π¦ Γ ππ¦/ππ₯=2β2π₯ ππ¦/ππ₯=(2 β 2π₯)/2π¦ ππ¦/ππ₯=(2 (1 β π₯))/2π¦ ππ¦/ππ₯=(1 β π₯)/π¦ Now if line is parallel to π₯βππ₯ππ Angle with π₯βππ₯ππ =0 π=0 Slope of π₯βππ₯ππ =tanβ‘π=tanβ‘0Β°=0 Now Slope of tangent = Slope of π₯βππ₯ππ ππ¦/ππ₯=0 (1 β π₯)/π¦=0 This will be possible only if Numerator is 0 i.e. 1βπ₯=0 π₯=1 Finding y when π₯=1 π₯^2+π¦^2β2π₯β3=0 (1)^2+π¦^2β2(1)β3=0 1+π¦^2β2β3=0 π¦^2β4=0 π¦^2=4 π¦=Β±β4 π¦=Β±2 Hence the point at which the tangent are parallel to the π₯βππ₯ππ are (π , π) & (π , βπ)

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Ex 6.3,19 You are here

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Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.