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Finding point when tangent is parallel/ perpendicular
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Finding point when tangent is parallel/ perpendicular
Last updated at May 29, 2023 by Teachoo
Question 15 Find the equation of the tangent line to the curve π¦=π₯2 β2π₯+7 which is : (a) parallel to the line 2π₯βπ¦+9=0We know that Slope of tangent is ππ¦/ππ₯ π¦=π₯2 β2π₯+7 Differentiating w.r.t.π₯ ππ¦/ππ₯=2π₯β2 Finding Slope of line 2π₯βπ¦+9=0 2π₯βπ¦+9=0 π¦=2π₯+9 π¦=2π₯+9 The Above Equation is of form π¦=ππ₯+π where m is Slope of line Hence, Slope of line 2π₯βπ¦+9 is 2 Now, Given tangent is parallel to 2π₯βπ¦+9=0 Slope of tangent = Slope of line 2π₯βπ¦+9 = 0 ππ¦/ππ₯=2 2π₯β2=2 2(π₯β1)=2 π₯=2 Finding y when π₯=2 , π¦=π₯^2β2π₯+7= (2)^2β2(2)+7=4β4+7=7 We need to find Equation of tangent passes through (2, 7) & Slope is 2 Equation of tangent is (π¦β7)=2(π₯β2) π¦β7=2π₯β4 π¦β2π₯β7+4=0 π¦β2π₯β3=0 Hence Required Equation of tangent is πβππβπ=π We know that Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Question 15 Find the equation of the tangent line to the curve π¦=π₯2β2π₯+7 which is (b) perpendicular to the line 5π¦β15π₯=13We know that Slope of tangent is ππ¦/ππ₯ π¦=π₯2 β2π₯+7 Differentiating w.r.t.π₯ ππ¦/ππ₯=2π₯β2 Finding Slope of line 5π¦β15π₯=13 5π¦β15π₯=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¦=ππ₯+π , where m is Slope of a line β΄ Slope = 3 Now, Given tangent is perpendicular to 5π¦β15π₯=13 Slope of tangent Γ Slope of line = β1 ππ¦/ππ₯ Γ 3=β1 ππ¦/ππ₯=(β1)/( 3) 2π₯β2=(β1)/( 3) 2π₯=(β1)/( 3)+2 2π₯=(β1 + 6)/3 2π₯=5/3 π₯=5/6 Finding y when π₯=5/6 π¦=π₯^2β2π₯+7=(5/6)^2β2(5/6)+7=25/36β10/6+7=217/36 β΄ Point is (5/6 ,217/36) Equation of tangent passing through (5/6 ,217/36) & having Slope (β1)/( 3) (π¦β217/36)=(β1)/( 3) (π₯β5/6) (36π¦ β217)/36=(β1)/( 3) (π₯β5/6) 36π¦ β217=(β36)/( 3) (π₯β5/( 6)) 36π¦ β217=β12(π₯β5/( 6)) 36π¦ β217=β12π₯+(12 Γ 5)/6 36π¦ β217=β12π₯+10 36π¦ β217=β12π₯+10 36π¦+12π₯β217β10=0 πππ+πππβπππ=π is Required Equation of tangent We know that Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1)