# Example 46

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 46 Find the equation of tangents to the curve y = cos (x + y), 2 x 2 that are parallel to the line x + 2y = 0. Given curve is = cos + , 2 2 We need to find equation of tangent which is parallel to the line + 2 = 0 We know that slope of tangent is = cos + Diff w.r.t. = + = sin + + = sin + + = sin + 1+ = sin + sin + . + sin + . = sin + 1+ + = + = + 1 + + Slope of tangent is + 1 + + Given line is + 2 = 0 2 = = 2 = 1 2 +0 The above equation is of the form = m + c where m is slope Slope of line is 1 2 We know that If two lines are parallel than their slopes are equal Given line + 2 = 0 is parallel to tangent Slope of tangent = Slope of line + 1 + + = 1 2 + 1 + + = 1 2 2 sin + =1+ + 2 sin + sin + =1 sin + =1 We know that sin 2 = 1 sin + = sin 2 Hence general solution of + is + = n + 1 n 2 Now, Finding points through which tangents pass. Given curve y = cos + Putting value of x + y y = cos + 1 2 y = 0 Putting y = 0 in x + = + 1 2 + 0 = n + 1 2 = n + 1 2 Since 2 2 Thus, = 3 2 & = 2 Points are 3 2 , 0 & 2 , 0 Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3 = 0 2x + 4y = 0 Example 46 Find the equation of tangents to the curve y = cos (x + y), 2 x 2 that are parallel to the line x + 2y = 0. Given curve is = cos + , 2 2 We need to find equation of tangent which is parallel to the line + 2 = 0 We know that slope of tangent is = cos + Diff w.r.t. = + = sin + + = sin + + = sin + 1+ = sin + sin + . + sin + . = sin + 1+ + = + = + 1 + + Slope of tangent is + 1 + + Given line is + 2 = 0 2 = = 2 = 1 2 +0 The above equation is of the form = m + c where m is slope Slope of line is 1 2 We know that If two lines are parallel than their slopes are equal Given line + 2 = 0 is parallel to tangent Slope of tangent = Slope of line + 1 + + = 1 2 + 1 + + = 1 2 2 sin + =1+ + 2 sin + sin + =1 sin + =1 We know that sin 2 = 1 sin + = sin 2 Hence general solution of + is + = n + 1 n 2 Now, Finding points through which tangents pass. Given curve y = cos + Putting value of x + y y = cos + 1 2 y = 0 Putting y = 0 in x + = + 1 2 + 0 = n + 1 2 = n + 1 2 Since 2 2 Thus, = 3 2 & = 2 Points are 3 2 , 0 & 2 , 0 Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3 = 0 2x + 4y = 0 Example 46 Find the equation of tangents to the curve y = cos (x + y), 2 x 2 that are parallel to the line x + 2y = 0. Given curve is = cos + , 2 2 We need to find equation of tangent which is parallel to the line + 2 = 0 We know that slope of tangent is = cos + Diff w.r.t. = + = sin + + = sin + + = sin + 1+ = sin + sin + . + sin + . = sin + 1+ + = + = + 1 + + Slope of tangent is + 1 + + Given line is + 2 = 0 2 = = 2 = 1 2 +0 The above equation is of the form = m + c where m is slope Slope of line is 1 2 We know that If two lines are parallel than their slopes are equal Given line + 2 = 0 is parallel to tangent Slope of tangent = Slope of line + 1 + + = 1 2 + 1 + + = 1 2 2 sin + =1+ + 2 sin + sin + =1 sin + =1 We know that sin 2 = 1 sin + = sin 2 Hence general solution of + is + = n + 1 n 2 Now, Finding points through which tangents pass. Given curve y = cos + Putting value of x + y y = cos + 1 2 y = 0 Putting y = 0 in x + = + 1 2 + 0 = n + 1 2 = n + 1 2 Since 2 2 Thus, = 3 2 & = 2 Points are 3 2 , 0 & 2 , 0 Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3 = 0 2x + 4y = 0

Chapter 6 Class 12 Application of Derivatives

Ex 6.3,5
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Ex 6.3,7 Important

Ex 6.3,12 Important

Ex 6.3,15 Important

Ex 6.3,26 Important

Example 35 Important

Example 37 Important

Example 38 Important

Example 40 Important

Ex 6.5,1 Important

Ex 6.5,5 Important

Ex 6.5,7 Important

Ex 6.5,11 Important

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Ex 6.5,23 Important

Ex 6.5,26 Important

Ex 6.5,28 Important

Example 46 Important You are here

Example 47 Important

Misc 6 Important

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Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.