Question:

Find the equation of all lines having slope 2 which are tangents to the curve y=\(\frac{1}{x-3},\) x≠3.

Updated On: Oct 19, 2023

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Solution and Explanation

The equation of the given curve is y=\(\frac{1}{x-3},\) x≠3.

The slope of the tangent to the given curve at any point (x, y) is given by,

\(\frac{dy}{dx}\)=\(-\frac{1}{(x-3)^2}\)

If the slope of the tangent is 2, then we have:

\(-\frac{1}{(x-3)^2}\)= 2

2(x-3)²=-1

(x-3)²=\(-\frac12\)

This is not possible since the L.HS. is positive while the R.H.S. is negative.

Hence, there is no tangent to the given curve having slope 2.

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A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got

m×n = -1

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