The normal to the rectangular hyperbola $xy=c^2$ at the point t meets the curve again at the point $t_1$ such that $t^3t_1$ is:
Answer & explanation
Correct answer: option 3
Consider general point $(ct,\frac{c}{t})$ on curve $xy = c^2$
⇒ Eq. of normal at $(ct,\frac{c}{t})≡xt^3-yt+c-ct^4=0$
It passes through $⇒(ct',\frac{c}{t'})≡ct'.t^3-\frac{c}{t'}.t+c-ct^4=0$
$⇒(t^3t'+1)(1-\frac{t}{t'})=0⇒t^3.t'=-1$