Examples
Last updated at December 16, 2024 by Teachoo
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Question 14 Find the equation of tangents to the curve y = cos (x + y), ā 2š ⤠x ⤠2š that are parallel to the line x + 2y = 0. Given curve is š¦ = cos (š„+š¦) We need to find equation of tangent which is parallel to the line š„ + 2š¦ = 0 We know that slope of tangent is šš¦/šš„ š¦ = cos (š„+š¦) Diff w.r.t. š šš¦/šš„ = š(ššš (š„ + š¦))/šš„ šš¦/šš„ = āsin (š„+š¦) š(š„ + š¦)/šš„ šš¦/šš„ = ā sin (š„+š¦) (šš„/šš„+šš¦/šš„) šš¦/šš„ = ā sin (š„+š¦) (1+šš¦/šš„) šš¦/šš„ = ā sin (š„+š¦) ā sin(š„+š¦). šš¦/šš„ šš¦/šš„ + sin (š„+š¦).šš¦/šš„ = ā sin (š„+š¦) šš¦/šš„ (1+š šš(š„+š¦))=āš šš(š„+š¦) š š/š š = (āššš(š + š))/(š + ššš( š + š) ) ā“ Slope of tangent is (āš šš(š„ + š¦))/(1 + š šš(š„ + š¦) ) Given line is š„ + 2š¦ = 0 2š¦ = āš„ š¦ = (āš„)/2 š = (( āš)/š)š+š The above equation is of the form š¦= mš„ + c where m is slope ā“ Slope of line is (ā1)/2 We know that If two lines are parallel than their slopes are equal Since line is parallel to tangent ā“ Slope of tangent = Slope of line (āššš(š + š))/(š + ššš(š + š) )= (āš)/š š šš(š„ + š¦)/(1 + š šš(š„ + š¦) )= 1/2 2 sin(š„+š¦)=1+š šš(š„+š¦) 2 sin (š„+š¦) ā sin(š„+š¦)=1 sin (š+š)=š Since sin š/2 = 1 sin(š„+š¦) = sin š /š Hence, (š„ + š¦) = nĻ + (ā1)^n š/2 Now, Finding points through which tangents pass Given curve y = cos (š„+š¦) Putting value of x + y y = cos (šš+(ā1)^š š/2) y = 0 Putting y = 0 in x š„ + š¦ = (šš+(ā1)^š š/2) š„ + 0 = nĻ + (ā1)^š š/2 š„ = nĻ + (ā1)^(š ) š/2 Since ā2Ļ ā¤ š„ ⤠2Ļ Thus, š„ = (ā3š)/2 & š„ = š/2 ā“ Points are ((āšš )/š , š) & (š /š , š) Putting n = 0 š„ = 0(š)+(ā1)^0 š/2 š„ = 0 + (š/2) š = š /š Putting n = ā1 š„ = ā1(š)+(ā1)^(ā1) š/2 š„ = āšāš/2 š„ = (ā2š ā š)/2 š = (āšš )/š Finding equation of tangents We know that Equation of line at (š„1 ,š¦1) & having slope at š is (š¦āš¦1)=š(š„āš„1) Equation of tangent at ((āšš )/š , š) & having slope (āš)/š is (š¦ā0) = (ā1)/2 (š„ā((ā3š)/2)) y = (ā1)/2 (š„+3š/2) y = (ā1)/2 ((2š„ + 3š)/2) 2x + 4y + 3Ļ = 0 Equation of tangent at (š /š , š) & having slope (āš)/š is (š¦ā0)= (ā1)/2 (š„āš/2) š¦ = (ā1)/2 ((2š„ ā š)/2) š¦ = (ā1)/4 (2š„āš) 4y = ā(2x ā Ļ) 2x + 4y ā Ļ = 0 Hence Required Equation of tangent are 2x + 4y + 3Ļ = 0 2x + 4y ā Ļ = 0