The slope of a tangent drawn at the point P(a | vedclass

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(B) Given the curve $y = \frac{1}{2x-5}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{(2x-5)^2} 	imes 2 = -\frac{2}{(2

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$3$

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(B) Given the curve $y = \frac{1}{2x-5}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{(2x-5)^2} 	imes 2 = -\frac{2}{(2x-5)^2}$.The slope of the tangent at $P(\alpha, \beta)$ is given as $-2$.So,$-\frac{2}{(2\alpha-5)^2} = -2$.$(2\alpha-5)^2 = 1$.$2\alpha-5 = 1$ or $2\alpha-5 = -1$.If $2\alpha-5 = 1$,then $2\alpha = 6$,so $\alpha = 3$. Then $\beta = \frac{1}{2(3)-5} = 1$. Point $P(3, 1)$ is in the first quadrant.If $2\alpha-5 = -1$,then $2\alpha = 4$,so $\alpha = 2$. Then $\beta = \frac{1}{2(2)-5} = -1$. Point $P(2, -1)$ is in the fourth quadrant.Since $P$ lies in the fourth quadrant,we have $\alpha = 2$ and $\beta = -1$.Therefore,$\alpha - \beta = 2 - (-1) = 3$.

The slope of a tangent drawn at the point $P(\alpha, \beta)$ lying on the curve $y=\frac{1}{2x-5}$ is $-2$. If $P$ lies in the fourth quadrant,then $\alpha-\beta=$

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