Ex 6.3,12 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 14, 2021 by Teachoo
Finding equation of tangent/normal when slope and curve are given
Finding equation of tangent/normal when slope and curve are given
Last updated at April 14, 2021 by Teachoo
Ex 6.3, 12 Find the equations of all lines having slope 0 which are tangent to the curve π¦ = 1/(π₯2 β2π₯ + 3) Given Curve is π¦ = 1/(π₯2 β2π₯ + 3) Slope of tangent is ππ¦/ππ₯ ππ¦/ππ₯=π(1/(π₯2 β 2π₯ + 3))/ππ₯ ππ¦/ππ₯=(π(π₯2 β 2π₯ + 3)^(β1))/ππ₯ ππ¦/ππ₯=β1(π₯2β2π₯+3)^(β2) . π(π₯^2β 2π₯ + 3)/ππ₯ ππ¦/ππ₯=β(π₯2β2π₯+3)^(β2) (2π₯β2) ππ¦/ππ₯=β2(π₯2β2π₯+3)^(β2) (π₯β1) ππ¦/ππ₯=(β2(π₯ β 1))/(π₯2 β 2π₯ + 3)^2 Hence Slope of tangent is (β2(π₯ β 1))/(π₯2 β 2π₯ + 3)^2 Given Slope of tangent is 0 β ππ¦/ππ₯=0 (β2(π₯ β 1))/(π₯2 β 2π₯ + 3)^2 =0 β2(π₯ β 1)=0 Γ(π₯2 β 2π₯ + 3)^2 β2(π₯ β 1)=0 (π₯β1)=0 π₯=1 Finding y when π₯=1 π¦=1/(π₯2 β 2π₯ + 3) π¦=1/((1)^2 β 2(1) + 3) π¦=1/(1 β 2 + 3) π¦=1/(2 ) Point is (π , π/π) Thus , tangent passes through (1 , 1/2) Equation of tangent at (1 , 1/2) & having Slope zero is (π¦ β1/2)=0(π₯β1) π¦ β1/2=0 π¦=1/2 Hence Equation of tangent is π=π/π We know that Equation of time passing through (π₯ , π¦) & having Slope m is (π¦ βπ¦1)=π(π₯ βπ₯2)