Finding equation of tangent/normal when point and curve is given

Chapter 6 Class 12 Application of Derivatives
Concept wise

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Question 6 Find the equations of the tangent and normal to the curve π₯^(2/3) + π¦^(2/3) = 2 at (1, 1).Given curve π₯^(2/3) + π¦^(2/3) = 2 Differentiating both sides w.r.t x 2/3 π₯^(1 β 2/3)+2/3 π¦^(1 β 2/3) ππ¦/ππ₯ = 0 2/3 π₯^((β1)/3)+2/3 π¦^((β1)/3) ππ¦/ππ₯ = 0 2/3 π¦^((β1)/3) ππ¦/ππ₯ = (β2)/3 π₯^((β1)/3) 1/π¦^(1/3) ππ¦/ππ₯ = (β1)/π₯^(1/3) ππ/ππ = β (π/π)^(π/π) Thus, Slope of tangent to the curve = β (π¦/π₯)^(1/3) At point (1, 1) Slope = β (π/π)^(π/π) = β1 Hence, Equation of tangent at point (1, 1) and with slope β1 is π¦β1=β1 (π₯β1) π¦β1=βπ₯+1 π+πβπ = π Also, Slope of Normal = (β1)/(πππππ ππ π‘ππππππ‘) = (β1)/(β1) = 1 Thus, Equation of normal at point (1, 1) and with slope 1 is π¦ β 1 = 1 (π₯ β 1) π¦ β 1 = π₯ β 1 π¦ =π₯ π βπ=π