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Find the equation of the normal to the c...

Find the equation of the normal to the curve `sqrt(x)-sqrt(y)=1` at (9, 4).

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`sqrt(x)-sqrt(y)=1`
Differentiating both sides w.r.t. x, we get,
`(1)/(2sqrt(x))-(1)/(2sqrt(y)).(dy)/(dx)=0`
`:.-(1)/(2sqrt(y)).(dy)/(dx)=-(1)/(2sqrt(x))`
`:.(dy)/(dx)=sqrt((y)/(x))`
`((dy)/(dx))_("at P"(9, 4))=sqrt((4)/(9))=(2)/(3)`= slope of the tangent at P(9, 4)
`:.` the slope of the normal at P(9, 4)
`=(-1)/((dy)/(dx))_("at P"(9, 4))=(-1)/(((2)/(3)))=-(3)/(2)`
`:.` the equation of the normal at P(9, 4) is
`y-4=(-3)/(2)(x-9)`
`:. 2y-8=-3x+27" ":.3x+2y=35`.
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