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Ex 6.3

Ex 6.3, 1
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Ex 6.3, 14 (i) Deleted for CBSE Board 2023 Exams

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Last updated at March 16, 2023 by Teachoo

Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iv) π¦=π₯2 ππ‘ (0, 0) Given Curve is π¦=π₯^2 Differentiating w.r.t.π₯ ππ¦/ππ₯=2π₯ We know that Slope of tangent is ππ¦/ππ₯ Given point is (0 , 0) Slope of tangent at (0 , 0) γππ¦/ππ₯βγ_((0 , 0) )=2(0) =0 We know that Slope of tangent Γ Slope if Normal =β1 0 Γ Slope if Normal =β1 Slope if Normal =(β1)/( 0) Equation of tangent at (0 , 0) & having Slope zero is (π¦β0)=0(π₯β0) We know that Equation of line at (π₯1 , π¦1)& having Slope m is π¦βπ¦1=π(π₯βπ₯1) Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (0, 0) & Slope 0 is (π¦β0)=0(π₯β0) π¦β0=0 π=π Equation of Normal at (0, 0) & Slope (β1)/0 is (π¦β0)=1/0 (π₯β0) 0 Γ (π¦β0)=1(π₯β0) 0=π₯β0 π=π