Solve all your doubts with Teachoo Black (new monthly pack available now!)

Are you in **school**? Do you **love Teachoo?**

We would love to talk to you! Please fill this form so that we can contact you

Ex 6.3

Ex 6.3, 1
Deleted for CBSE Board 2023 Exams

Ex 6.3,2 Deleted for CBSE Board 2023 Exams

Ex 6.3,3 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,4 Deleted for CBSE Board 2023 Exams

Ex 6.3, 5 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,6 Deleted for CBSE Board 2023 Exams

Ex 6.3,7 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,8 Deleted for CBSE Board 2023 Exams

Ex 6.3,9 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,10 Deleted for CBSE Board 2023 Exams

Ex 6.3,11 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,12 Deleted for CBSE Board 2023 Exams

Ex 6.3,13 Deleted for CBSE Board 2023 Exams

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

Ex 6.3, 14 (ii) Important Deleted for CBSE Board 2023 Exams

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

Ex 6.3, 14 (iv) Important Deleted for CBSE Board 2023 Exams

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

Ex 6.3,15 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,16 Deleted for CBSE Board 2023 Exams

Ex 6.3,17 Deleted for CBSE Board 2023 Exams You are here

Ex 6.3,18 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,19 Deleted for CBSE Board 2023 Exams

Ex 6.3,20 Deleted for CBSE Board 2023 Exams

Ex 6.3,21 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,22 Deleted for CBSE Board 2023 Exams

Ex 6.3,23 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,24 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,25 Deleted for CBSE Board 2023 Exams

Ex 6.3,26 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.3,27 (MCQ) Deleted for CBSE Board 2023 Exams

Last updated at April 14, 2021 by Teachoo

Ex 6.3, 17 Find the points on the curve 𝑦=𝑥3 at which the slope of the tangent is equal to the y-coordinate of the pointLet the Point be (ℎ , 𝑘) on the Curve 𝑦=𝑥3 Where Slope of tangent at (ℎ , 𝑘)=𝑦−𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 (ℎ, 𝑘) i.e. 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=𝑘 Given 𝑦=𝑥^3 Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=3𝑥^2 ∴ Slope of tangent at (ℎ , 𝑘) is 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=3ℎ^2 From (1) 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=𝑘 3ℎ^2=𝑘 Also Point (ℎ , 𝑘) is on the Curve 𝑦=𝑥^3 Point (ℎ , 𝑘) must Satisfy the Equation of Curve i.e. 𝑘=ℎ^3 Now our equations are 3ℎ^2=𝑘 …(1) & 𝑘=ℎ^3 …(2) Putting Value of 𝑘=3ℎ^2 in (3) 3ℎ^2=ℎ^3 ℎ^3−3ℎ^2=0 ℎ^2 (ℎ−3)=0 ℎ^2=0 ℎ=0 ℎ−3=0 ℎ=3 When 𝒉=𝟎 3ℎ^2=𝑘 3(0)=𝑘 𝑘=0 Hence, point is (0, 0) When 𝒉=𝟑 3ℎ^2=𝑘 3(3)^2=𝑘 𝑘=27 Hence, point is (3 , 27)