Question

If the normal at $(ct,\frac{c}{t})$ on the curve $xy=c^2$ meets the curve again in $t^{\prime }$, then

• A. $t^{\prime }=-\frac{1}{t^3}$
• B. $t^{\prime }=\frac{-1}{t}$
• C. $t^{\prime }=\frac{1}{t^2}$
• D. $t^{\prime 2}=\frac{-1}{t^2}$

Answer 1

The correct option is A $t^{\prime }=-\frac{1}{t^3}$

$xy=c^2$ and point (ct, c/t)

Let us equation of the normal

$y=\frac{c^2}{x}$

$\frac{dy}{dx}=\frac{-c^2}{x^2}$

hence slope of normal is given by $\frac{x^2}{c^2}$ & (ct , c/t) is

$t^2$ and its equation is

$y=t^2(x-ct)+c/t$

$\therefore x{t}^3-yt-c{t}^4+c=0$

At this passes through $(c{t}^1,c/t^1)$

$c{t}^1.t^3-\frac{c}{t^1}.t-c{t}^4+c=0$

$c{t}^3.(t^1-t)+\frac{c}{t^1}(t^1-t)=0$

$(c{t}^3+\frac{c}{t^1})(t^1-t)=0$

$c{t}^3+\frac{c}{t^1}=0$ $\to c{t}^3=\frac{c}{t^1}$

$t^3=\frac{-1}{t^1}$

$\to t^1=\frac{-1}{t^3}$