# Example 46 - Chapter 6 Class 12 Application of Derivatives

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

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Chapter 6 Class 12 Application of Derivatives

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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.