Let a curve y = f(x), x ∈ (0, ∞) pass through the points P(1,...

Given,

A curve $y = f\left(x\right), x \in \left(0, \infty\right)$ pass through the points $P\left(1, \dfrac{3}{2}\right)$ and $Q\left(a, \dfrac{1}{2}\right)$,

And the tangent at any point $R ( b, f ( b ) )$ to the given curve cuts the $y$-axis at the point $S ( 0, c )$ such that $bc = 3,$

Now tangent to the curve is given by,

$Y - y = m\left(X - x\right), m = \dfrac{dy}{dx}$

Now at $Y -$axis, put $X = 0$ we get,

$Y = y - mx$ or $c = y - mx$ as point $S\left(0, c\right)$ given,

Now given $bc = 3$

$⇒ x\left(y - mx\right) = 3$ {here $b = x$ on the curve}

$⇒ y - \dfrac{xdy}{dx} = \dfrac{3}{x}$

$⇒ \dfrac{ydx - xdy}{{x}^{2}} = \dfrac{3dx}{x} \cdot \dfrac{1}{{x}^{2}}$

$⇒ d\left(\dfrac{- y}{x}\right) = 3d\left(\dfrac{{x}^{- 2}}{- 2}\right)$

Now integrating both side we get,

$⇒ \dfrac{y}{x} = \dfrac{3}{2{x}^{2}} + C$,

Now given curve passes through $P\left(1, \dfrac{3}{2}\right)$, so $C = 0$

So, equation of curve will be $2xy = 3$

Now curve also passes through the point $Q\left(a, \dfrac{1}{2}\right)$

So, $\alpha = 3$

Hence, $P\left(1, \dfrac{3}{2}\right), Q\left(3, \dfrac{1}{2}\right)$

So, by distance formula we get, ${\left(PQ\right)}^{2} = 4 + 1 = 5$