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

Question

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

Professor PC Thomas Classes · Answer

(5) : Equation of tangent at $R(b, f(b))$ is $y-f(b)=f^{\prime}(b)(x-b)$\ Now, it passes through $S(0, c)$\ $	herefore \quad c-f(b)=f^{\prime}(b)(0-b)$\ $\Rightarrow \quad \frac{3}{b}-f(b)=-b f^{\prime}(b)$\ $\Rightarrow b f^{\prime}(b)-f(b)=-\frac{3}{b}$\ $\Rightarrow \quad \frac{b f^{\prime}(b)-f(b)}{b^{2}}=-\frac{3}{b^{3}}$\ $\Rightarrow d\left(\frac{f(b)}{b}ight)=-\frac{3}{b^{3}} \Rightarrow \frac{f(b)}{b}=\frac{3}{2 b^{2}}+M$\ Now, this curve passes through $P\left(1, \frac{3}{2}ight)$.\ $	herefore \quad \frac{3}{2}=\frac{3}{2}+M \quad$ So, $f(b)=\frac{3}{2 b}$\ Also, it passes through $Q\left(a, \frac{1}{2}ight)$\ $\Rightarrow \frac{1}{2}=\frac{3}{2 a} \Rightarrow a=3 \Rightarrow Q 
eq\left(3, \frac{1}{2}ight)$\ $	herefore \quad(P Q)^{2}=2^{2}+1^{2}=5$\