Let a curve y=f(x),x (0,) pass through the points P(1,(3)/(2)) and Q(a,(1)/(2)).

Question

Let a curve $y=f(x),x\in (0,\infty)$ pass through the points $P(1,\frac{3}{2})$ and $Q(a,\frac{1}{2})$. If 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 $bc=3$, then $(PQ{)}^{2}$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Equation of tangent at $R(b,f(b))$

$y-f(b)={f}^{'}(b)(x-b)$

which passes through $S(0,c)$

$\begin{gathered} 	herefore c-f(b)={f}^{'}(b)(0-b) \ b{f}^{'}(b)-f(b)=-c \ \Rightarrow b{f}^{'}(b)-f(b)=\frac{-3}{b}(\because bc=3) \ \Rightarrow \frac{b{f}^{'}(b)-f(b)}{{b}^{2}}=\frac{-3}{{b}^{3}} \ \Rightarrow d(\frac{f(b)}{b})=\frac{-3}{{b}^{3}} \ \Rightarrow \frac{f(b)}{b}=\frac{3}{2{b}^{2}}+c \end{gathered}$

which passes through $P(1,\frac{3}{2})$

$\begin{gathered} \Rightarrow \frac{3/2}{1}=\frac{3}{2}+c \ \Rightarrow c=0 \ 	herefore f(b)=\frac{3}{2{b}^{2}}	imes b \ \Rightarrow f(b)=\frac{3}{2b} \end{gathered}$

$\because $ It passes through $Q(a,\frac{1}{2})$

$\begin{gathered} 	herefore \frac{1}{2}=\frac{3}{2a} \ \Rightarrow a=3 \ 	herefore P\equiv (1,\frac{3}{2})Q\equiv (3,\frac{1}{2}) \ 	herefore (PQ{)}^{2}=(3-1{)}^{2}+{(\frac{1}{2}-\frac{3}{2})}^{2}=4+1=5 \end{gathered}$

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