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Show that the curves `x = y^2` and `xy = k` cut at right angles; if `8k^2 = 1` |
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Answer» Let the point of intersection of the given curves be `(x_(1), y_(1))` Then, `y^(2) = x hArr 2y + (dy)/(dx) = 1 hArr ((dy)/(dx))_((x_(1)"," y_(1))) = (1)/(2y_(1)) hArr m_(1) = (1)/(2y_(1))` `xy = k hArr x (dy)/(dx) + y = 0 hArr (dy)//(dx) = (-y)/(x) hArr ((dy)/(dx))_((x_(1)"," y_(1))) = m_(2) = (-y_(1))/(x_(1))` `:. m_(1) m_(2) = -1 hArr (1)/(2y_(1)) xx (-y_(1))/(x_(1)) = -1 hArr x_(1) = (1)/(2)` `(x_(1), y_(1))` lies on `y^(2) = x hArr y_(1)^(2) = x_(1) hArr x_(1) = (1)/(2)` `(x_(1), y_(1))` lies on `xy = k hArr x_(1) y_(1) = k hArr x_(1)^(2) y_(1)^(2) = k^(2)` `hArr k^(2) = ((1)/(4) xx (1)/(2)) = (1)/(8)` `hArr 8k^(2) = 1` |
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