







Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Examples
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
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 You are here
Example 47 Important
Example 48 Important
Example 49
Example 50 Important
Example 51
Last updated at April 19, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Example 46 Find the equation of tangents to the curve y = cos (x + y), β 2π β€ x β€ 2π that are parallel to the line x + 2y = 0. Given curve is π¦ = cos (π₯+π¦) We need to find equation of tangent which is parallel to the line π₯ + 2π¦ = 0 We know that slope of tangent is ππ¦/ππ₯ π¦ = cos (π₯+π¦) Diff w.r.t. π ππ¦/ππ₯ = π(πππ (π₯ + π¦))/ππ₯ ππ¦/ππ₯ = βsin (π₯+π¦) π(π₯ + π¦)/ππ₯ ππ¦/ππ₯ = β sin (π₯+π¦) (ππ₯/ππ₯+ππ¦/ππ₯) ππ¦/ππ₯ = β sin (π₯+π¦) (1+ππ¦/ππ₯) ππ¦/ππ₯ = β sin (π₯+π¦) β sin(π₯+π¦). ππ¦/ππ₯ ππ¦/ππ₯ + sin (π₯+π¦).ππ¦/ππ₯ = β sin (π₯+π¦) ππ¦/ππ₯ (1+π ππ(π₯+π¦))=βπ ππ(π₯+π¦) π π/π π = (βπππ(π + π))/(π + πππ( π + π) ) β΄ Slope of tangent is (βπ ππ(π₯ + π¦))/(1 + π ππ(π₯ + π¦) ) Given line is π₯ + 2π¦ = 0 2π¦ = βπ₯ π¦ = (βπ₯)/2 π = (( βπ)/π)π+π The above equation is of the form π¦= mπ₯ + c where m is slope β΄ Slope of line is (β1)/2 We know that If two lines are parallel than their slopes are equal Since line is parallel to tangent β΄ Slope of tangent = Slope of line (βπππ(π + π))/(π + πππ(π + π) )= (βπ)/π π ππ(π₯ + π¦)/(1 + π ππ(π₯ + π¦) )= 1/2 2 sin(π₯+π¦)=1+π ππ(π₯+π¦) 2 sin (π₯+π¦) β sin(π₯+π¦)=1 sin (π+π)=π Since sin π/2 = 1 sin(π₯+π¦) = sin π /π Hence, (π₯ + π¦) = nΟ + (β1)^n π/2 Now, Finding points through which tangents pass Given curve y = cos (π₯+π¦) Putting value of x + y y = cos (ππ+(β1)^π π/2) y = 0 Putting y = 0 in x π₯ + π¦ = (ππ+(β1)^π π/2) π₯ + 0 = nΟ + (β1)^π π/2 π₯ = nΟ + (β1)^(π ) π/2 Since β2Ο β€ π₯ β€ 2Ο Thus, π₯ = (β3π)/2 & π₯ = π/2 β΄ Points are ((βππ )/π , π) & (π /π , π) Putting n = 0 π₯ = 0(π)+(β1)^0 π/2 π₯ = 0 + (π/2) π = π /π Putting n = β1 π₯ = β1(π)+(β1)^(β1) π/2 π₯ = βπβπ/2 π₯ = (β2π β π)/2 π = (βππ )/π Finding equation of tangents We know that Equation of line at (π₯1 ,π¦1) & having slope at π is (π¦βπ¦1)=π(π₯βπ₯1) Equation of tangent at ((βππ )/π , π) & having slope (βπ)/π is (π¦β0) = (β1)/2 (π₯β((β3π)/2)) y = (β1)/2 (π₯+3π/2) y = (β1)/2 ((2π₯ + 3π)/2) 2x + 4y + 3Ο = 0 Equation of tangent at (π /π , π) & having slope (βπ)/π is (π¦β0)= (β1)/2 (π₯βπ/2) π¦ = (β1)/2 ((2π₯ β π)/2) π¦ = (β1)/4 (2π₯βπ) 4y = β(2x β Ο) 2x + 4y β Ο = 0 Hence Required Equation of tangent are 2x + 4y + 3Ο = 0 2x + 4y β Ο = 0