Misc 20 - Slope of tangent to x = t2 + 3t - 8, y = 2t2 - 2t - 5 - Miscellaneous

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Misc 20 The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2,– 1) is (A) ﷐22﷮7﷯ (B) ﷐6﷮7﷯ (C) ﷐7﷮6﷯ (D) ﷐− 6﷮7﷯ We need to find slope of tangent at (2, -1) We know that slope of tangent is ﷐𝑑𝑦﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯= ﷐﷐𝑑𝑦﷮𝑑𝑡﷯﷮﷐𝑑𝑥﷮𝑑𝑡﷯﷯ Thus, ﷐𝑑𝑦﷮𝑑𝑥﷯= ﷐﷐𝑑𝑦﷮𝑑𝑡﷯﷮﷐𝑑𝑥﷮𝑑𝑡﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯= ﷐4𝑡 − 2﷮2𝑡 + 3﷯ Now we need to find value of t Given point is (2, –1) Putting 𝑥 = 2 & 𝑦 = –1 in the curve Since t = 2 is common in both parts So, we will calculate ﷐𝑑𝑦﷮𝑑𝑥﷯= ﷐4𝑡 −2﷮2𝑡 −3﷯ at t = 2 At t = 2 ﷐𝑑𝑦﷮𝑑𝑥﷯= ﷐4𝑡 −2﷮2𝑡 + 3﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯= ﷐4 ﷐2﷯ − 2﷮2 ﷐2﷯ + 3﷯= ﷐8 − 2﷮4 +3﷯= ﷐6﷮7﷯ Hence the correct answer is (B)

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