Question

The length of the chord along the normal at $\left(ct,\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$t$}\right.\right)$ on the curve $xy=c^2$ is

• A. $\frac{c}{t^3}{\left(1+t^4\right)}^{\frac{3}{2}}$
• B. $-\frac{c}{t^3}{\left(1+t^4\right)}^{\frac{3}{2}}$
• C. $-\frac{c}{t^3}{\left(1-t^4\right)}^{\frac{3}{2}}$
• D. None of these

Answer 1

The correct option is A

$\frac{c}{t^3}{\left(1+t^4\right)}^{\frac{3}{2}}$

Explanation for correct option:

Given, $xy=c^2$ and point $\left(ct,\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$t$}\right.\right)$

The slope of the normal is

$$\begin{gathered} y=\frac{c^2}{x} \\ \Rightarrow \frac{dy}{dx}=\frac{-c^2}{x^2} \\ \Rightarrow {\left[\frac{dy}{dx}\right]}_{\left(ct,\frac{c}{t}\right)}=-\frac{1}{t^2} \end{gathered}$$

Therefore the equation of the normal at point $\left(ct,\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$t$}\right.\right)$ is

$$\begin{gathered} \left(y-\frac{c}{t}\right)=-\left(\frac{1}{{\displaystyle \frac{-1}{t^2}}}\right)\left(x-ct\right) \\ \Rightarrow y=t^2\left(x-ct\right)+\frac{c}{t} \\ \therefore x{t}^3-yt-c{t}^4+c=0 \end{gathered}$$

At this passes through $\left(c{t}_1,\frac{c}{t_1}\right)$

$$\begin{gathered} \Rightarrow c{t}_1·t^3-\frac{c}{t_1}·t-c{t}^4+c=0 \\ \Rightarrow c{t}^3·\left(t_1-t\right)+\frac{c}{t_1}\left(t_1-t\right)=0 \\ \Rightarrow \left(t_1-t\right)\left(c{t}^3+\frac{c}{t_1}\right)=0 \\ \therefore c{t}^3+\frac{c}{t_1}=0 \\ \Rightarrow \frac{1}{t_1}=-t^3 \\ \Rightarrow t_1=-\frac{1}{t^3} \end{gathered}$$

Normal at $\left(ct,\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$t$}\right.\right)$ meets the curve at $Q\left(\frac{-c}{t^3},-c{t}^3\right)$

$$\begin{gathered} PQ=\sqrt{{\left(ct+\frac{c}{t^3}\right)}^2+{\left(\frac{c}{t}+c{t}^3\right)}^2} \\ =c\left(t^4+1\right)\sqrt{\frac{1}{t^6}+\frac{1}{t^2}} \\ =\left(\frac{c}{t^3}\right){\left(t^4+1\right)}^{\frac{3}{2}} \end{gathered}$$

Hence, option A is correct.