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The curves `x=y^(2) and xy =a^(3)` cut orthogonally at a point, then a =A. `(1)/(3)`B. 3C. 2D. `(1)/(2)` |
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Answer» Correct Answer - D We have, `x=y^(2) " " (i) and , xy =a^(3) " " …(ii) ` These two curves intersect at `P(a, a^(2))`. Now, `x=y^(2) rArr (dy)/(dx)=(1)/(2y) rArr m_(1) = ((dy)/(dx))_(p) = (1)/(2a^(2))` and, `xy=a rArr x(dy)/(dx)+y=0 rArr (dy)/(dx) = (y)/(x) rArr m_(2) = ((dy)/(dx))_(p) = -a` If the two curves intersect orthogonally, then `m_(1)m_(2) = -1 rArr (1)/(2a^(2)) xx -a = -1 rArr a = (1)/(2)` |
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