Ex 6.3,18 - Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.3,18 For the curve 𝑦=4𝑥3 −2𝑥5, find all the points at which the tangent passes through the origin. Let ﷐ℎ , 𝑘﷯ be the Required Point on the Curve at which tangent is to be taken Given Curve is 𝑦=4﷐𝑥﷮3﷯−2﷐𝑥﷮5﷯ Since Point ﷐ℎ , 𝑘﷯ is on the Curve ⇒ ﷐ℎ , 𝑘﷯ will satisfy the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘=4﷐ℎ﷮3﷯−2﷐ℎ﷮5﷯ We know that Slope of tangent to the Curve is ﷐𝑑𝑦﷮𝑑𝑥﷯ 𝑦=4﷐𝑥﷮3﷯−2﷐𝑥﷮5﷯ Differentiating w.r.t. 𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐𝑑﷐4﷐𝑥﷮3 ﷯− 2﷐𝑥﷮5﷯﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=12﷐𝑥﷮2﷯−10﷐𝑥﷮4﷯ Since tangent is taken from ﷐ℎ , 𝑘﷯ Slope of tangent at ﷐ℎ, 𝑘﷯ is ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮﷐ℎ, 𝑘﷯﷯=12﷐ℎ﷮2﷯−10﷐ℎ﷮4﷯ Now, Equation of tangent at ﷐ℎ , 𝑘﷯ & having Slope 12﷐ℎ﷮2﷯−10﷐ℎ﷮4﷯ is ﷐𝑦−𝑘﷯=12﷐ℎ﷮2﷯−10﷐ℎ﷮4﷯﷐𝑥−ℎ﷯ Also, Given tangent passes through Origin ⇒ ﷐0 , 0﷯ will satisfies the Equation of tangent Putting x = 0, y = 0 in equation ﷐0 −𝑘﷯=12﷐ℎ﷮2﷯−10﷐ℎ﷮4﷯﷐0−ℎ﷯ −𝑘=12﷐ℎ﷮2﷯−10﷐ℎ﷮4﷯﷐−ℎ﷯ −𝑘=−12﷐ℎ﷮3﷯+10﷐ℎ﷮5﷯ Now our Equations are 𝑘=4﷐ℎ﷮3﷯−2﷐ℎ﷮5﷯ …(1) −𝑘=−12﷐ℎ﷮3﷯+10﷐ℎ﷮5﷯ …(2) Adding (1) and (2) 𝑘−𝑘=4﷐ℎ﷮3﷯−2﷐ℎ﷮5﷯+﷐−12﷐ℎ﷮3﷯+10﷐ℎ﷮5﷯﷯ 0=4﷐ℎ﷮3﷯−12﷐ℎ﷮3﷯−2﷐ℎ﷮5﷯+10﷐ℎ﷮5﷯ 0=− 8﷐ℎ﷮3﷯+8﷐ℎ﷮5﷯ − 8﷐ℎ﷮3﷯+8﷐ℎ﷮5﷯=0 − 8﷐ℎ﷮3﷯﷐1−﷐ℎ﷮2﷯﷯=0 Hence the Required point on the Curve are ﷐0 , 0﷯, ﷐1 , 2﷯ & ﷐−1 , −2﷯