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Find the equation of all lines having slope 2 which are tangents to the curve `y=1/(x-3), x!=3`

Answer» Slope of tangent = 2
Equation of curve ` y = 1/(x-3)`
`rArr (dy)/(dx) = (-1)/((x-3)^(2))`
Slope of tangent at point `(x, y) = (-1)/((x-3)^(2))`
`:. (-1)/((x-3)^(2)) = 2`
` rArr (x-3)^(2) = - 1/2`
which is impossible.
Therefore, there is no tangent of the given curve with slope 2.


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