1.

Find the equation of all lines having slope 2 and being tangent to the curve `y+2/(x-3)=0`.

Answer» Eqaution of the curve is,
`y +2/(x-3) = 0`
Differentiating it w.r.t. `x`,
`=>dy/dx - 2/(x-3)^2 = 0`
`=> dy/dx = 2/(x-3)^2`
Now, it is given that slope of the tangent is `2`.
`:. dy/dx = 2`
`=>2/(x-3)^2 = 2`
`=>(x-3)^2 = 1`
`=>x-3 = +-1`
`=>x = 4 and x = 2`
When `x = 4`,
`y+2/(4-3) = 0`
`=>y = -2`
So, equation of tangent with this point,`y+2 = 2(x-4) => 2x - y = 10`
When `x = 2`,
`y+2/(2-3) = 0`
`=>y = 2`
So, equation of tangent with this point,`y-2 = 2(x-2) => 2x - y = 2.`


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