Solve all your doubts with Teachoo Black (new monthly pack available now!)
Example 17 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 19, 2021 by Teachoo
Examples
Example 1
Example 2
Example 3
Example 4 Important
Example 5
Example 6
Example 7
Example 8 Important
Example 9 Important
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14 Deleted for CBSE Board 2023 Exams
Example 15 Deleted for CBSE Board 2023 Exams
Example 16 Deleted for CBSE Board 2023 Exams
Example 17 Important Deleted for CBSE Board 2023 Exams You are here
Example 18 Deleted for CBSE Board 2023 Exams
Example 19 Deleted for CBSE Board 2023 Exams
Example 20 Deleted for CBSE Board 2023 Exams
Example 21 Deleted for CBSE Board 2023 Exams
Example 22 Deleted for CBSE Board 2023 Exams
Example 23 Deleted for CBSE Board 2023 Exams
Example 24
Example 25
Example 26
Example 27
Example 28 Important
Example 29
Example 30 Important
Example 31
Example 32 Important
Example 33 Important
Example 34
Example 35 Important
Example 36
Example 37 Important
Example 38 Important
Example 39
Example 40 Important
Example 41 Important
Example 42 Important
Example 43 Important
Example 44 Important
Example 45 Important Deleted for CBSE Board 2023 Exams
Example 46 Important Deleted for CBSE Board 2023 Exams
Example 47 Important
Example 48 Important
Example 49
Example 50 Important
Example 51
Chapter 6 Class 12 Application of Derivatives
Serial order wise
Transcript
Example 17 Find points on the curve π₯^2/4 + π¦^2/25 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.Given curve π₯^2/4 + π¦^2/25 = 1 Slope of the tangent is ππ¦/ππ₯ Finding π π/π π 2π₯/4+(2π¦ )/25 Γ ππ¦/ππ₯= 0 π₯/2 + 2π¦/25 ππ¦/ππ₯ = 0 2π¦/25 ππ¦/ππ₯ = (βπ₯)/2 ππ¦/ππ₯ = (βπ₯)/2 Γ 25/2π¦ π π/π π = (βπππ)/ππ Hence, ππ¦/ππ₯ = β (β25π₯)/4π¦ (i) Tangent is parallel to x-axis If the tangent is parallel to x-axis, its slope is 0 Hence, ππ¦/ππ₯ = 0 (β25π₯)/4π¦ = 0 x = 0 Putting x = 0 in equation of the curve π₯^2/4 + π¦^2/25 = 1 0/4 + π¦^2/25 = 1 π¦^2 = 25 y = Β± 5 Hence, the points are (0, 5) and (0, β5) (ii) Tangent is parallel to y-axis If the tangent is parallel to y-axis, its slope is 1/0 Hence, ππ¦/ππ₯ = 1/0 (β25π₯)/4π¦ = 1/0 y = 0 Putting y = 0 in equation of the curve π₯^2/4 + π¦^2/25 = 1 π₯^2/4 + 0/25 = 1 π₯^2/4 = 1 x2 = 4 x = Β± 2 Hence, the points are (2, 0) and (β2, 0)
