Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

### Transcript

Ex 6.3, 17 Find the points on the curve π¦=π₯3 at which the slope of the tangent is equal to the y-coordinate of the pointLet the Point be (β , π) on the Curve π¦=π₯3 Where Slope of tangent at (β , π)=π¦βπππππππππ‘π ππ (β, π) i.e. γππ¦/ππ₯βγ_((β, π) )=π Given π¦=π₯^3 Differentiating w.r.t.π₯ ππ¦/ππ₯=3π₯^2 β΄ Slope of tangent at (β , π) is γππ¦/ππ₯βγ_((β, π) )=3β^2 From (1) γππ¦/ππ₯βγ_((β, π) )=π 3β^2=π Also Point (β , π) is on the Curve π¦=π₯^3 Point (β , π) must Satisfy the Equation of Curve i.e. π=β^3 Now our equations are 3β^2=π β¦(1) & π=β^3 β¦(2) Putting Value of π=3β^2 in (3) 3β^2=β^3 β^3β3β^2=0 β^2 (ββ3)=0 β^2=0 β=0 ββ3=0 β=3 When π=π 3β^2=π 3(0)=π π=0 Hence, point is (0, 0) When π=π 3β^2=π 3(3)^2=π π=27 Hence, point is (3 , 27)