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if the straight line `xcosalpha+y sin alpha=p` touches the curve `x^my^n=a^(m+n)` prove that `p^(m+n)m^mn^n=(m+n)^(m+n) a^(m+n) sin^nalpha cos^malpha` |
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Answer» Let the point of contact be `(x_(1), y_(1))` The equation of the given curve is `x^(m) y^(n) = a^(m + n)` This, on differentiation, gives `mx^(m-1) y^(n) + x^(m).ny^(n-1) .(dy)/(dx) = 0 rArr (dy)/(dx) = (-my)/(nx)` `:. ((dy)/(dx))_((x_(1)"," y_(1))) = (-my_(1))/(nx_(1))` So, the equation of the tangent at `(x_(1), y_(1)) " is " (y - y_(1))/(x - x_(1)) = (-my_(1))/(nx_(1))` i.e., `my_(1) x + nx_(1) y = (m + n) x_(1)y_(1)`....(i) This is identical with `c cos alpha + y sin alpha = p`...(ii) Comparing coefficients, we get `(my_(1))/(cos alpha) = (nx_(1))/(sin alpha) = ((m + n) x_(1)y_(1))/(p)` `:. x_(1) = (pm)/((m +n) cos alpha) and y_(1) = (pn)/((m + n) sin alpha)` Since `(x_(1)y_(1))` lies on the given curve, `x_(1)^(m) y_(1)^(n) = a^(m +n)` `:. {("pm")/((m + n) cos alpha)}^(m).{("pm")/((m +n) sin alpha)}^(n) = a^(m +n)` or `p^(m +n) .m^(m). n^(n) = (m +n)^(m + n) .a^(m +n) cos^(m) alpha sin^(n) alpha` |
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