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Ex 6.3
Ex 6.3,2
Ex 6.3,3 Important
Ex 6.3,4
Ex 6.3, 5 Important
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
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Ex 6.3,13
Ex 6.3, 14 (i)
Ex 6.3, 14 (ii) Important
Ex 6.3, 14 (iii)
Ex 6.3, 14 (iv) Important
Ex 6.3, 14 (v)
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 You are here
Ex 6.3,20
Ex 6.3,21 Important
Ex 6.3,22
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25
Ex 6.3,26 (MCQ) Important
Ex 6.3,27 (MCQ)
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)