Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

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Question 15 Find the equation of the tangent line to the curve π¦=π₯2 β2π₯+7 which is : (a) parallel to the line 2π₯βπ¦+9=0We know that Slope of tangent is ππ¦/ππ₯ π¦=π₯2 β2π₯+7 Differentiating w.r.t.π₯ ππ¦/ππ₯=2π₯β2 Finding Slope of line 2π₯βπ¦+9=0 2π₯βπ¦+9=0 π¦=2π₯+9 π¦=2π₯+9 The Above Equation is of form π¦=ππ₯+π where m is Slope of line Hence, Slope of line 2π₯βπ¦+9 is 2 Now, Given tangent is parallel to 2π₯βπ¦+9=0 Slope of tangent = Slope of line 2π₯βπ¦+9 = 0 ππ¦/ππ₯=2 2π₯β2=2 2(π₯β1)=2 π₯=2 Finding y when π₯=2 , π¦=π₯^2β2π₯+7= (2)^2β2(2)+7=4β4+7=7 We need to find Equation of tangent passes through (2, 7) & Slope is 2 Equation of tangent is (π¦β7)=2(π₯β2) π¦β7=2π₯β4 π¦β2π₯β7+4=0 π¦β2π₯β3=0 Hence Required Equation of tangent is πβππβπ=π We know that Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Question 15 Find the equation of the tangent line to the curve π¦=π₯2β2π₯+7 which is (b) perpendicular to the line 5π¦β15π₯=13We know that Slope of tangent is ππ¦/ππ₯ π¦=π₯2 β2π₯+7 Differentiating w.r.t.π₯ ππ¦/ππ₯=2π₯β2 Finding Slope of line 5π¦β15π₯=13 5π¦β15π₯=13 5π¦=15π₯+13 π¦=1/5 (15π₯+13) π¦=15/5 π₯+13/5 π¦=3π₯+13/5 Above Equation is of form π¦=ππ₯+π , where m is Slope of a line β΄ Slope = 3 Now, Given tangent is perpendicular to 5π¦β15π₯=13 Slope of tangent Γ Slope of line = β1 ππ¦/ππ₯ Γ 3=β1 ππ¦/ππ₯=(β1)/( 3) 2π₯β2=(β1)/( 3) 2π₯=(β1)/( 3)+2 2π₯=(β1 + 6)/3 2π₯=5/3 π₯=5/6 Finding y when π₯=5/6 π¦=π₯^2β2π₯+7=(5/6)^2β2(5/6)+7=25/36β10/6+7=217/36 β΄ Point is (5/6 ,217/36) Equation of tangent passing through (5/6 ,217/36) & having Slope (β1)/( 3) (π¦β217/36)=(β1)/( 3) (π₯β5/6) (36π¦ β217)/36=(β1)/( 3) (π₯β5/6) 36π¦ β217=(β36)/( 3) (π₯β5/( 6)) 36π¦ β217=β12(π₯β5/( 6)) 36π¦ β217=β12π₯+(12 Γ 5)/6 36π¦ β217=β12π₯+10 36π¦ β217=β12π₯+10 36π¦+12π₯β217β10=0 πππ+πππβπππ=π is Required Equation of tangent We know that Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1)

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.