# Example 46

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 46 Find the equation of tangents to the curve y = cos (x + y), – 2𝜋 ≤ x ≤ 2 𝜋 that are parallel to the line x + 2y = 0. Given curve is 𝑦 = cos 𝑥+𝑦 , –2π ≤ 𝑥 ≤ 2π We need to find equation of tangent which is parallel to the line 𝑥 + 2𝑦 = 0 We know that slope of tangent is 𝑑𝑦𝑑𝑥 𝑦 = cos 𝑥+𝑦 Diff w.r.t. 𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑐𝑜𝑠𝑥 + 𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = –sin 𝑥+𝑦 𝑑𝑥 + 𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 𝑑𝑥𝑑𝑥+𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 1+𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 – sin𝑥+𝑦. 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 + sin 𝑥+𝑦.𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 𝑑𝑦𝑑𝑥1+𝑠𝑖𝑛𝑥+𝑦=−𝑠𝑖𝑛𝑥+𝑦 𝑑𝑦𝑑𝑥 = −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛 𝑥 + 𝑦 ∴ Slope of tangent is −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦 Given line is 𝑥 + 2𝑦 = 0 2𝑦 = –𝑥 𝑦 = −𝑥2 𝑦 = −12𝑥+0 The above equation is of the form 𝑦= m𝑥 + c where m is slope ⇒ Slope of line is −12 We know that If two lines are parallel than their slopes are equal Given line 𝑥 + 2𝑦 = 0 is parallel to tangent ∴ Slope of tangent = Slope of line −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦= −12 𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦= 12 2 sin𝑥+𝑦=1+𝑠𝑖𝑛𝑥+𝑦 2 sin 𝑥+𝑦 – sin𝑥+𝑦=1 sin 𝑥+𝑦=1 We know that sin 𝜋2 = 1 sin𝑥+𝑦 = sin 𝜋2 Hence general solution of 𝑥 + 𝑦 is 𝑥 + 𝑦 = nπ + −1n 𝜋2 Now, Finding points through which tangents pass. Given curve y = cos 𝑥+𝑦 Putting value of x + y y = cos 𝑛𝜋+−1𝑛𝜋2 y = 0 Putting y = 0 in x 𝑥 + 𝑦 = 𝑛𝜋+−1𝑛𝜋2 𝑥 + 0 = nπ + −1𝑛𝜋2 𝑥 = nπ + −1𝑛 𝜋2 Since 2π ≤ 𝑥 ≤ 2π Thus, 𝑥 = −3𝜋2 & 𝑥 = 𝜋2 ∴ Points are −3𝜋2 , 0 & 𝜋2 , 0 Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3π = 0 2x + 4y – π = 0 Example 46 Find the equation of tangents to the curve y = cos (x + y), – 2𝜋 ≤ x ≤ 2 𝜋 that are parallel to the line x + 2y = 0. Given curve is 𝑦 = cos 𝑥+𝑦 , –2π ≤ 𝑥 ≤ 2π We need to find equation of tangent which is parallel to the line 𝑥 + 2𝑦 = 0 We know that slope of tangent is 𝑑𝑦𝑑𝑥 𝑦 = cos 𝑥+𝑦 Diff w.r.t. 𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑐𝑜𝑠𝑥 + 𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = –sin 𝑥+𝑦 𝑑𝑥 + 𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 𝑑𝑥𝑑𝑥+𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 1+𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 – sin𝑥+𝑦. 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 + sin 𝑥+𝑦.𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 𝑑𝑦𝑑𝑥1+𝑠𝑖𝑛𝑥+𝑦=−𝑠𝑖𝑛𝑥+𝑦 𝑑𝑦𝑑𝑥 = −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛 𝑥 + 𝑦 ∴ Slope of tangent is −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦 Given line is 𝑥 + 2𝑦 = 0 2𝑦 = –𝑥 𝑦 = −𝑥2 𝑦 = −12𝑥+0 The above equation is of the form 𝑦= m𝑥 + c where m is slope ⇒ Slope of line is −12 We know that If two lines are parallel than their slopes are equal Given line 𝑥 + 2𝑦 = 0 is parallel to tangent ∴ Slope of tangent = Slope of line −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦= −12 𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦= 12 2 sin𝑥+𝑦=1+𝑠𝑖𝑛𝑥+𝑦 2 sin 𝑥+𝑦 – sin𝑥+𝑦=1 sin 𝑥+𝑦=1 We know that sin 𝜋2 = 1 sin𝑥+𝑦 = sin 𝜋2 Hence general solution of 𝑥 + 𝑦 is 𝑥 + 𝑦 = nπ + −1n 𝜋2 Now, Finding points through which tangents pass. Given curve y = cos 𝑥+𝑦 Putting value of x + y y = cos 𝑛𝜋+−1𝑛𝜋2 y = 0 Putting y = 0 in x 𝑥 + 𝑦 = 𝑛𝜋+−1𝑛𝜋2 𝑥 + 0 = nπ + −1𝑛𝜋2 𝑥 = nπ + −1𝑛 𝜋2 Since 2π ≤ 𝑥 ≤ 2π Thus, 𝑥 = −3𝜋2 & 𝑥 = 𝜋2 ∴ Points are −3𝜋2 , 0 & 𝜋2 , 0 Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3π = 0 2x + 4y – π = 0 Example 46 Find the equation of tangents to the curve y = cos (x + y), – 2𝜋 ≤ x ≤ 2 𝜋 that are parallel to the line x + 2y = 0. Given curve is 𝑦 = cos 𝑥+𝑦 , –2π ≤ 𝑥 ≤ 2π We need to find equation of tangent which is parallel to the line 𝑥 + 2𝑦 = 0 We know that slope of tangent is 𝑑𝑦𝑑𝑥 𝑦 = cos 𝑥+𝑦 Diff w.r.t. 𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑐𝑜𝑠𝑥 + 𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = –sin 𝑥+𝑦 𝑑𝑥 + 𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 𝑑𝑥𝑑𝑥+𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 1+𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 – sin𝑥+𝑦. 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 + sin 𝑥+𝑦.𝑑𝑦𝑑𝑥 = – sin 𝑥+𝑦 𝑑𝑦𝑑𝑥1+𝑠𝑖𝑛𝑥+𝑦=−𝑠𝑖𝑛𝑥+𝑦 𝑑𝑦𝑑𝑥 = −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛 𝑥 + 𝑦 ∴ Slope of tangent is −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦 Given line is 𝑥 + 2𝑦 = 0 2𝑦 = –𝑥 𝑦 = −𝑥2 𝑦 = −12𝑥+0 The above equation is of the form 𝑦= m𝑥 + c where m is slope ⇒ Slope of line is −12 We know that If two lines are parallel than their slopes are equal Given line 𝑥 + 2𝑦 = 0 is parallel to tangent ∴ Slope of tangent = Slope of line −𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦= −12 𝑠𝑖𝑛𝑥 + 𝑦1 + 𝑠𝑖𝑛𝑥 + 𝑦= 12 2 sin𝑥+𝑦=1+𝑠𝑖𝑛𝑥+𝑦 2 sin 𝑥+𝑦 – sin𝑥+𝑦=1 sin 𝑥+𝑦=1 We know that sin 𝜋2 = 1 sin𝑥+𝑦 = sin 𝜋2 Hence general solution of 𝑥 + 𝑦 is 𝑥 + 𝑦 = nπ + −1n 𝜋2 Now, Finding points through which tangents pass. Given curve y = cos 𝑥+𝑦 Putting value of x + y y = cos 𝑛𝜋+−1𝑛𝜋2 y = 0 Putting y = 0 in x 𝑥 + 𝑦 = 𝑛𝜋+−1𝑛𝜋2 𝑥 + 0 = nπ + −1𝑛𝜋2 𝑥 = nπ + −1𝑛 𝜋2 Since 2π ≤ 𝑥 ≤ 2π Thus, 𝑥 = −3𝜋2 & 𝑥 = 𝜋2 ∴ Points are −3𝜋2 , 0 & 𝜋2 , 0 Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3π = 0 2x + 4y – π = 0

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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

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