1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.3

Transcript

Ex 6.3,19 Find the points on the curve π₯^2+π¦^2 β2π₯ β3=0 at which the tangents are parallel to the π₯βππ₯ππ  Given tangent is parallel to the π₯βππ₯ππ  β Slope of tangent = Slope of π₯βππ₯ππ  We know that Slope of tangent is ππ¦/ππ₯ π₯^2+π¦^2 β2π₯ β3=0 Differentiating w.r.t.π₯ π(π₯^2 + π¦^2 β2π₯ β3)/ππ₯=0 π(π₯^2 )/ππ₯+π(π¦^2 )/ππ₯βπ(2π₯)/ππ₯βπ(3)/ππ₯=0 2π₯+π(π¦^2 )/ππ₯ Γ ππ¦/ππ¦β2β0=0 2π₯+π(π¦^2 )/ππ₯ Γ ππ¦/ππ¦β2=0 π(π¦^2 )/ππ₯ Γ ππ¦/ππ₯=2β2π₯ 2π¦ Γ ππ¦/ππ₯=2β2π₯ ππ¦/ππ₯=(2 β 2π₯)/2π¦ ππ¦/ππ₯=(2 (1 β π₯))/2π¦ ππ¦/ππ₯=(1 β π₯)/π¦ Now if line is parallel to π₯βππ₯ππ  Angle with π₯βππ₯ππ =0 π=0 Slope of π₯βππ₯ππ =tanβ‘π=tanβ‘0Β°=0 Now Slope of tangent = Slope of π₯βππ₯ππ  ππ¦/ππ₯=0 (1 β π₯)/π¦=0 This will be possible only if Numerator is 0 i.e. 1βπ₯=0 π₯=1 Finding y when π₯=1 π₯^2+π¦^2β2π₯β3=0 (1)^2+π¦^2β2(1)β3=0 1+π¦^2β2β3=0 π¦^2β4=0 π¦^2=4 π¦=Β±β4 π¦=Β±2 Hence the point at which the tangent are parallel to the π₯βππ₯ππ  are (π , π) & (π , βπ)

Ex 6.3