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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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