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Find the equation of tangent to the curv...

Find the equation of tangent to the curve xy = 16 which passes through (0,2) .

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Given function is
xy = 16
Differentiating with respect to x
`x(dy)/(dx) + y = 0`
`implies (dy)/(dx) = (-(y)/(x))`
Let point of contact is (h,k)
`(dy)/(dx)|_("(h,k)") = (-(k)/(h))`
Equation of tangent
`(y-k) = -((k)/(h)) (x - h)`
This passes through (0,2)
So , `2 - k = - (k)/(h) xx (-h)`
`implies 2- k = k`
`implies k = 1`
Substituting k = 1 in hk = 16
`implies h = 16`
So equation of tangent will be
`y - 1 = - (1)/(16) (x - 16)`
`implies 16 y - 16 = - x + 16`
`implies x = 16y = 32`
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