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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Example 17 Find points on the curve ๐‘ฅ^2/4 + ๐‘ฆ^2/25 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. Given curve ๐‘ฅ^2/4 + ๐‘ฆ^2/25 = 1 Slope of the tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Finding ๐’…๐’š/๐’…๐’™ 2๐‘ฅ/4+(2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)/25 = 0 ๐‘ฅ/2 + 2๐‘ฆ/25 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 2๐‘ฆ/25 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’๐‘ฅ)/2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’๐‘ฅ)/2 ร— 25/2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’25๐‘ฅ)/4๐‘ฆ Hence, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’ (โˆ’25๐‘ฅ)/4๐‘ฆ (i) Tangent is parallel to x-axis If the tangent is parallel to x โ€“ axis, its slope is 0 Hence, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 (โˆ’25๐‘ฅ)/4๐‘ฆ = 0 x = 0 Putting x = 0 in equation of the curve ๐‘ฅ^2/4 + ๐‘ฆ^2/25 = 1 0/4 + ๐‘ฆ^2/25 = 1 ๐‘ฆ^2 = 25 y = ยฑ 5 Hence, the points are (0, 5) and (0, โˆ’5) (ii) Tangent is parallel to y-axis If the tangent is parallel to y โˆ’ axis, its slope is 1/0 Hence, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/0 (โˆ’25๐‘ฅ)/4๐‘ฆ = 1/0 y = 0 Putting y = 0 in equation of the curve ๐‘ฅ^2/4 + ๐‘ฆ^2/25 = 1 ๐‘ฅ^2/4 + 0/25 = 1 ๐‘ฅ^2/4 = 1 x2 = 4 x = ยฑ 2 Hence, the points are (2, 0) and (โˆ’2, 0)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.