Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.3,15 Find the equation of the tangent line to the curve = 2 2 +7 which is : (a) parallel to the line 2 +9=0 We know that Slope of tangent is = 2 2 +7 Differentiating w.r.t. =2 2 Finding Slope of line 2 +9=0 2 +9=0 =2 +9 =2 +9 The Above Equation is of the form = + where m is Slope of line Hence Slope of line 2 +9 is 2 Now, Given tangent is parallel to 2 +9=0 Slope of tangent = Slope of line 2 +9 = 0 =2 2 2=2 2 1 =2 =2 Finding y when =2 , = 2 2 +7= 2 2 2 2 +7=4 4+7=7 We need to find Equation of tangent passes through 2 , 7 & Slope is 2 Equation of tangent is 7 =2 2 7=2 4 2 7+4=0 2 3=0 Hence Required Equation of tangent parallel to = Ex 6.3,15 Find the equation of the tangent line to the curve = 2 2 +7 which is (b) perpendicular to the line 5 15 =13 We know that Slope of tangent is = 2 2 +7 Differentiating w.r.t. =2 2 Finding Slope of line 5 15 =13 5 15 =13 5 =15 +13 = 1 5 15 +13 = 15 5 + 13 5 =3 + 13 5 Above Equation is of the form = + , where m is Slope of a line Slope = 3 Now, Given tangent is perpendicular to 5 15 =13 Slope of tangent Slope of line = 1 3= 1 = 1 3 2 2= 1 3 2 = 1 3 +2 2 = 1 + 6 3 2 = 5 3 = 5 6 Finding y when = 5 6 = 2 2 +7= 5 6 2 2 5 6 +7= 25 36 10 6 +7= 217 36 Point is 5 6 , 217 36 Equation of tangent passing through 5 6 , 217 36 & having Slope 1 3 217 36 = 1 3 5 6 36 217 36 = 1 3 5 6 36 217= 36 3 5 6 36 217= 12 5 6 36 217= 12 + 12 5 6 36 217= 12 +10 36 217= 12 +10 36 +12 217 10=0 + = is Required Equation of tangent