If a chord AB of the curve $y=\frac{3x-2}{x-2}$ is a normal to the curve at the point A,  subtending an angle $\theta$  at the point P(2, 3). Then the value of  $\frac{\sin \left(\theta -\overline{)PAB}\right)}{\sin \left(\theta +\overline{)PAB}\right)}$   is equal to

$y=\frac{3x-2}{x-2}$

$\left(x-2\right)\left(y-3\right)=4$

Shift the curve so that centre is origin; XY = 4

A(2t₁ 2/t₁)

Normal at A(t₁) meets the curve again at B(t₂)

$\because \left.\begin{array}{cc}{t}_2=\frac{1}{t_1^3},& m\left(oA\right)=\frac{1}{t_1^2}\\ & m\left(oB\right)=\frac{1}{t_2^2}\end{array}\right\}$

$$\begin{gathered} \tan \theta =\frac{{\displaystyle \frac{1}{{t_1}^2}}-\frac{1}{{t_2}^2}}{1+\frac{1}{{t_1}^2}\frac{1}{{t_2}^2}}=\frac{\frac{1}{{t_1}^2}-{t_1}^6}{1+\frac{1}{{t_1}^2}\left({t_1}^6\right)}=\frac{1-{t_1}^8}{{t_1}^2\left(1+{t_1}^4\right)}= \\ \tan \theta =\frac{1-t_1^4}{t_1^2} \end{gathered}$$

$also,\tan \left(\overline{)OAB}\right)=\frac{t_1^4-1}{2t_1^2}$

$\frac{\tan \theta }{\tan A}=-2$

$\frac{\tan \theta -\tan A}{\tan \theta +\tan A}=\frac{-2-1}{-2+1}=3$

$\frac{\sin \left(\theta -\overline{)PAB}\right)}{\sin \left(\theta +\overline{)PAB}\right)}$=3