Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Question 14 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 ππ¦/ππ₯ π¦ = cos (π₯+π¦) Diff w.r.t. π ππ¦/ππ₯ = π(πππ (π₯ + π¦))/ππ₯ ππ¦/ππ₯ = βsin (π₯+π¦) π(π₯ + π¦)/ππ₯ ππ¦/ππ₯ = β sin (π₯+π¦) (ππ₯/ππ₯+ππ¦/ππ₯) ππ¦/ππ₯ = β sin (π₯+π¦) (1+ππ¦/ππ₯) ππ¦/ππ₯ = β sin (π₯+π¦) β sin(π₯+π¦). ππ¦/ππ₯ ππ¦/ππ₯ + sin (π₯+π¦).ππ¦/ππ₯ = β sin (π₯+π¦) ππ¦/ππ₯ (1+π ππ(π₯+π¦))=βπ ππ(π₯+π¦) ππ/ππ = (βπππ(π + π))/(π + πππ( π + π) ) β΄ Slope of tangent is (βπ ππ(π₯ + π¦))/(1 + π ππ(π₯ + π¦) ) Given line is π₯ + 2π¦ = 0 2π¦ = βπ₯ π¦ = (βπ₯)/2 π = (( βπ)/π)π+π The above equation is of the form π¦= mπ₯ + c where m is slope β΄ Slope of line is (β1)/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 (βπππ(π + π))/(π + πππ(π + π) )= (βπ)/π π ππ(π₯ + π¦)/(1 + π ππ(π₯ + π¦) )= 1/2 2 sin(π₯+π¦)=1+π ππ(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 + (β1)^n π/2 Now, Finding points through which tangents pass Given curve y = cos (π₯+π¦) Putting value of x + y y = cos (ππ+(β1)^π π/2) y = 0 Putting y = 0 in x π₯ + π¦ = (ππ+(β1)^π π/2) π₯ + 0 = nΟ + (β1)^π π/2 π₯ = nΟ + (β1)^(π ) π/2 Since β2Ο β€ π₯ β€ 2Ο Thus, π₯ = (β3π)/2 & π₯ = π/2 β΄ Points are ((βππ)/π , π) & (π/π , π) Putting n = 0 π₯ = 0(π)+(β1)^0 π/2 π₯ = 0 + (π/2) π = π/π Putting n = β1 π₯ = β1(π)+(β1)^(β1) π/2 π₯ = βπβπ/2 π₯ = (β2π β π)/2 π = (βππ)/π Finding equation of tangents We know that Equation of line at (π₯1 ,π¦1) & having slope at π is (π¦βπ¦1)=π(π₯βπ₯1) Equation of tangent at ((βππ)/π , π) & having slope (βπ)/π is (π¦β0) = (β1)/2 (π₯β((β3π)/2)) y = (β1)/2 (π₯+3π/2) y = (β1)/2 ((2π₯ + 3π)/2) 2x + 4y + 3Ο = 0 Equation of tangent at (π/π , π) & having slope (βπ)/π is (π¦β0)= (β1)/2 (π₯βπ/2) π¦ = (β1)/2 ((2π₯ β π)/2) π¦ = (β1)/4 (2π₯βπ) 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