Question

# Find the equation for all  lines having slopes $$2$$ and being tangent to the curve  $$y + \frac{2}{{x - 3}} = 0$$.

Solution

## Consider the given equation, $$y+\dfrac{2}{x-3}=0$$    …..(1) Differentiate with respect to x , we get   $$\dfrac{dy}{dx}+\dfrac{d}{dx}\left( \dfrac{2}{x-3} \right)=\dfrac{d}{dx}0$$  $$\dfrac{dy}{dx}+2\left( -1 \right)\times \dfrac{1}{{{\left( x-3 \right)}^{2}}}=0$$  $$\dfrac{dy}{dx}=\dfrac{2}{{{\left( x-3 \right)}^{2}}}$$ But given that $$\dfrac{dy}{dx}=2$$   $$2=\dfrac{2}{{{\left( x-3 \right)}^{2}}}$$  $$1=\dfrac{1}{{{\left( x-3 \right)}^{2}}}$$  $${{\left( x-3 \right)}^{2}}=1$$  $$x-3=\pm 1$$  $$x=4,x=2$$ When,$$x=4$$ put in equation 1st we get $$y=-2$$ When, $$x=2$$,then$$y=2$$ Now, if $$\left( x,y \right)=\left( 4,-2 \right)$$   \begin{align} y-\left( -2 \right)=2\left( x-4 \right)  y+2=2x-8  2x-y=10  Mathematics

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