# Ex 6.3,19 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 14, 2021 by Teachoo

Ex 6.3

Ex 6.3, 1
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Last updated at April 14, 2021 by Teachoo

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 that 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₯ Finding π π/π π π₯^2+π¦^2 β2π₯ β3=0 Given that 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₯ Finding π π/π π 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 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, Since 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 1βπ₯=0 Γ y 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 required points are (π , π) & (π , βπ)