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Ex 6.3, 21 Find the equation of the normal to the curve ๐ฆ=๐ฅ^3+2๐ฅ+6 which are parallel to the line ๐ฅ+14๐ฆ+4=0.Let (โ , ๐) be the point on the Curve at which Normal is to be taken Given Curve is ๐ฆ=๐ฅ^3+2๐ฅ+6 Since point (โ , ๐) is on the Curve โด (โ , ๐) will satisfies the Equation of Curve Putting ๐ฅ=โ , ๐ฆ=๐ ๐=โ^3+2โ+6 We know that Slope of a tangent to the Curve is ๐๐ฆ/๐๐ฅ ๐ฆ=๐ฅ^3+2๐ฅ+6 Differentiating w.r.t. ๐ฅ ๐๐ฆ/๐๐ฅ=3๐ฅ^2+2 Since tangent to be taken from (โ , ๐) Slope of tangent at (โ , ๐) is ใ๐๐ฆ/๐๐ฅโใ_((โ, ๐) )=3โ^2+2 We know that Slope of tangent ร Slope of Normal =โ1 (3โ^2+2) ร Slope of Normal =โ1 Slope of Normal = (โ1)/(3โ^2 + 2) Also, Given that Normal is parallel to the line ๐ฅ+14๐ฆ+4=0 If two lines are parallel then slopes are equal โ Slopes of Normal = Slope of line ๐ฅ+14๐ฆ+4=0 Now, line is ๐ฅ+14๐ฆ+4=0 14๐ฆ=โ๐ฅโ4 ๐ฆ=(โ ๐ฅ โ 4)/14 ๐ฆ=((โ1)/14)๐ฅโ(4/14) The above equation is of the form ๐ฆ=๐๐ฅ+๐ where m is slope โด Slope of line ๐ฅ+14๐ฆ+4=0 is (โ1)/14 Now, Slope of Normal = Slope of line ๐ฅ+14๐ฆ+4=0 (โ1)/(3โ^2 + 2)=(โ1)/( 14) 1/(3โ^2 + 2)=1/( 14) 14=3โ^2+2 3โ^2+2=14 3โ^2=14โ2 3โ^2=12 โ^2=12/3 โ^2=4 โ=ยฑโ4 โ=ยฑ2 When ๐=๐ ๐=โ^3+2โ+6 ๐=(2)^3+2(2)+6 ๐=8+4+6 ๐=18 โด Point is (๐ , ๐๐) When ๐=โ๐ ๐=โ^3+2โ+6 ๐=(โ2)^3+2(โ2)+6 ๐=โ8โ4+6 ๐=โ 6 โด Point is (โ๐, โ๐) Finding equation of normal We know that Equation of line at (๐ฅ1 , ๐ฆ1)& having Slope m is ๐ฆโ๐ฆ1=๐(๐ฅโ๐ฅ1) Equation of 1st Normal at (2 , 18) & having Slope (โ1)/14 is (๐ฆโ18)=(โ1)/14 (๐ฅโ2) 14(๐ฆโ18)=โ(๐ฅโ2) 14๐ฆโ252=โ๐ฅ+2 14๐ฆ+๐ฅโ252โ2=0 ๐+๐๐๐โ๐๐๐=๐ Equation of 2nd Normal at (โ2 , โ6) & having Slope (โ1)/14 is (๐ฆโ(โ6))=(โ1)/14 (๐ฅโ(โ2)) ๐ฆ+6=(โ1)/14 (๐ฅ+2) 14๐ฆ+84=โ๐ฅโ2 14๐ฆ+๐ฅ+84+2=0 ๐+๐๐๐+๐๐=๐
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