Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

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### Transcript

Ex 6.3, 9 Find the point on the curve π¦=π₯^3β11π₯+5 at which the tangent is π¦=π₯ β11.Equation of Curve is π¦=π₯^3β11π₯+5 We know that Slope of tangent is ππ¦/ππ₯ ππ¦/ππ₯=π(π₯^3 β 11π₯ + 5)/ππ₯ ππ¦/ππ₯=γ3π₯γ^2β11 Also, Given tangent is π¦=π₯β12 Comparing with π¦=ππ₯+π , when m is the Slope Slope of tangent =1 From (1) and (2) ππ¦/ππ₯=1 3π₯^2β11=1 3π₯^2=1+11 3π₯^2=12 π₯^2=12/3 π₯^2=4 π₯=Β±2 When π=π π¦=(2)^3β11(2)+5 π¦=8β22+5 π¦=β 9 So, Point is (2 , β9) When π=βπ π¦=(β2)^3β11(β2)+5 π¦=β 8+22+5 π¦=19 So, Point is (β2 , 19) Hence , Points (2 , β9) & (β2 , 19) But (β2, 19) does not satisfy line y = x β 11 As 19 β  β2 β 11 β΄ Only point is (2, β9)