Example 17 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 19, 2021 by Teachoo
Examples
Example 1
Deleted for CBSE Board 2022 Exams
Example 2 Deleted for CBSE Board 2022 Exams
Example 3 Deleted for CBSE Board 2022 Exams
Example 4 Important Deleted for CBSE Board 2022 Exams
Example 5 Deleted for CBSE Board 2022 Exams
Example 6 Deleted for CBSE Board 2022 Exams
Example 7
Example 8 Important
Example 9 Important
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15
Example 16
Example 17 Important You are here
Example 18
Example 19
Example 20
Example 21 Deleted for CBSE Board 2022 Exams
Example 22 Deleted for CBSE Board 2022 Exams
Example 23 Deleted for CBSE Board 2022 Exams
Example 24 Deleted for CBSE Board 2022 Exams
Example 25 Deleted for CBSE Board 2022 Exams
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 Deleted for CBSE Board 2022 Exams
Example 43 Important Deleted for CBSE Board 2022 Exams
Example 44 Important Deleted for CBSE Board 2022 Exams
Example 45 Important Deleted for CBSE Board 2022 Exams
Example 46 Important
Example 47 Important
Example 48 Important
Example 49 Deleted for CBSE Board 2022 Exams
Example 50 Important
Example 51
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)
