Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Example 17 - Chapter 6 Class 12 Application of Derivatives
Last updated at Jan. 7, 2020 by Teachoo
Transcript
Example 17 Find points on the curve ๐ฅ^2/4 + ๐ฆ^2/25 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. Given curve ๐ฅ^2/4 + ๐ฆ^2/25 = 1 Slope of the tangent is ๐๐ฆ/๐๐ฅ Finding ๐ ๐/๐ ๐ 2๐ฅ/4+(2๐ฆ ๐๐ฆ/๐๐ฅ)/25 = 0 ๐ฅ/2 + 2๐ฆ/25 ๐๐ฆ/๐๐ฅ = 0 2๐ฆ/25 ๐๐ฆ/๐๐ฅ = (โ๐ฅ)/2 ๐๐ฆ/๐๐ฅ = (โ๐ฅ)/2 ร 25/2๐ฆ ๐๐ฆ/๐๐ฅ = (โ25๐ฅ)/4๐ฆ Hence, ๐๐ฆ/๐๐ฅ = โ (โ25๐ฅ)/4๐ฆ (i) Tangent is parallel to x-axis If the tangent is parallel to x โ axis, its slope is 0 Hence, ๐๐ฆ/๐๐ฅ = 0 (โ25๐ฅ)/4๐ฆ = 0 x = 0 Putting x = 0 in equation of the curve ๐ฅ^2/4 + ๐ฆ^2/25 = 1 0/4 + ๐ฆ^2/25 = 1 ๐ฆ^2 = 25 y = ยฑ 5 Hence, the points are (0, 5) and (0, โ5) (ii) Tangent is parallel to y-axis If the tangent is parallel to y โ axis, its slope is 1/0 Hence, ๐๐ฆ/๐๐ฅ = 1/0 (โ25๐ฅ)/4๐ฆ = 1/0 y = 0 Putting y = 0 in equation of the curve ๐ฅ^2/4 + ๐ฆ^2/25 = 1 ๐ฅ^2/4 + 0/25 = 1 ๐ฅ^2/4 = 1 x2 = 4 x = ยฑ 2 Hence, the points are (2, 0) and (โ2, 0)
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Chapter 6 Class 12 Application of Derivatives
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About the Author
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.