# How do you find the equation of the tangent and normal line to the curve y=(2x+3)/(3x-2) at (1,5)?

Jan 14, 2017

$\left(1\right) \text{Eqn. of Tgt. : } 13 x + y - 18 = 0$.

$\left(2\right) \text{Eqn. of Normal : } x - 13 y + 64 = 0$.

#### Explanation:

Recall that $\frac{\mathrm{dy}}{\mathrm{dx}}$ gives the slope of tangent (tgt.) to the curve at

the general point $\left(x , y\right)$.

Now, $y = \frac{2 x + 3}{3 x - 2} = \frac{\frac{2}{3} \left(3 x - 2\right) + \frac{13}{3}}{3 x - 2} = \frac{2}{3} + \frac{13}{3} {\left(3 x - 2\right)}^{-} 1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 0 + \frac{13}{3} \left\{- 1 {\left(3 x - 2\right)}^{- 1 - 1} \frac{d}{\mathrm{dx}} \left(3 x - 2\right)\right\}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{13}{3 x - 2} ^ 2$.

$\therefore \text{ The Slope of the Tgt. at the Point "(1,5)" is, } - 13.$

Also, the tgt. passes through $\left(1 , 5\right)$.

Therefore, by the Slope-Pt. Form , the eqn. of tgt. is,

$y - 5 = - 13 \left(x - 1\right) , i . e . , 13 x + y - 18 = 0$.

As regards the eqn. of the Normal, it is $\bot$ to tgt. at $\left(1 , 5\right)$.

So, the eqn. of normal is $: y - 5 = \frac{1}{13} \left(x - 1\right) , \mathmr{and} , x - 13 y + 64 = 0$.

Enjoy Maths.!