Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

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### Transcript

Ex 6.3, 16 Show that the tangents to the curve π¦=7π₯3+11 at the points where π₯=2 and π₯ =β2 are parallel.We know that 2 lines are parallel y Slope of 1st line = Slope of 2nd line π1=π2 We know that Slope of tangent is ππ¦/ππ₯ Given Curve is π¦=7π₯^3+11 Differentiating w.r.t.π₯ ππ¦/ππ₯=π(7π₯3 + 11)/ππ₯ ππ¦/ππ₯=21π₯^2 We need to show that tangent at π₯=2 & tangent at π₯=β2 are parallel i.e. we need to show (Slope of tangent at π₯=2) = (Slope of tangent at π₯=β2) Now, Slope of tangent = ππ¦/ππ₯=21π₯^2 Slope of tangent at π₯=2 γππ¦/ππ₯βγ_(π₯ = 2)=21(2)^2=21 Γ4=84 & Slope of tangent at π₯=β2 γππ¦/ππ₯βγ_(π₯ =β 2)=21(β2)^2=21 Γ4=84 Since, (Slope of tangent at π₯=2) = (Slope of tangent at π₯=β2) Thus, tangent at π₯=2 & tangent at π₯ are parallel Hence Proved