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Prove that the curve `y = x^2` and `xy = k` intersect orthogonally if `8k^2 = 1`. |
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Answer» Given, equation of curves are `=2x=y^(2)` ………….(i) and `2xy=k`……………….(ii) `rArr y=k/(2x)` [From eq.(ii)] From Eq.(i), `2x=(k/(2x))^(2)` `rArr 8x^(3)=k^(2)` `rArr x^(3)=1/8k^(2)` `rArr x=1/2k^(2//3)` `rArr y=k/(2x) = k/(2.1/2k^(2//3)) = k^(1//3)` Thus, we get point of intersection of curves which is `(1/2k^(2//3), k^(1//3))` From Eqs. (i) and (ii), `2=2y(dy)/(dx)` and `2[x.(dy)/(dx) + y.1]=0` `rArr (dy)/(dx) =1/y` and `(dy)/(dx) =(-2y)/(2x) =-y/3` `rArr (dy)/(dx) = (-2y)/(2x)=-y/x` `rArr (dy)/(dx)_(1/2k^(2//3),k^(1//3)) = 1/k^(1//3)` [say `m_(1)`] and `(dx)/(dy)_(1/2k^(2//3),k^(1//3)) = k^(1//3)/(1/2k^(2//3)) = -2k^(-1//3)` [say `m_(2)`] Since, the curves intersect orthogonally. i.e., `m_(1).m_(2)=-1` `rArr 1/k^(1//3).(-2k^(-1//3))=-1` `rArr -2k^(-2//3) = -1` `rArr 2/(k^(2//3)) =1` `k^(2//3)=2` `therefore k^(2)=8` Which is the required condition. |
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