

Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 6.3,8 Find a point on the curve ๐ฆ=(๐ฅโ2)^2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4). Given Curves is ๐ฆ=(๐ฅโ2)^2 Let AB be the chord joining the Point (2 , 0) & (4 ,4) & CD be the tangent to the Curve ๐ฆ=(๐ฅโ2)^2 Given that tangent is Parallel to chord i.e. CD โฅ AB โ Slope of CD = Slope of AB Slope of tangent CD is ๐๐ฆ/๐๐ฅ ๐ฆ=(๐ฅโ2)^2 ๐๐ฆ/๐๐ฅ=(๐(๐ฅโ2)^2)/๐๐ฅ ๐๐ฆ/๐๐ฅ=2(๐ฅโ2) . (๐ฅโ2) ๐๐ฆ/๐๐ฅ=2(๐ฅโ2) . (1โ0) ๐๐ฆ/๐๐ฅ=2(๐ฅโ2) & Slope of AB As AB is chord joining Points (2 , 0) & (4 , 4) Slope of AB =(4 โ 0)/(4 โ 2) =4/2 =2 Now Slope of CD = Slope of AB 2(๐ฅโ2)=2 ๐ฅโ2=2/2 ๐ฅโ2=1 ๐ฅ=3 Finding y when ๐ฅ=3 ๐ฆ=(๐ฅโ2)^2 ๐ฆ=(3โ2)^2 ๐ฆ=(1)^2 ๐ฆ=1 Hence Point is (๐ , ๐) Hence the tangent is parallel to the chord at (3 ,1)
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