Question

The normal to the rectangular hyperbola xy = c² at the point 't' meets the curve again at a point 't' such that

• A. t³t' = -1
• B. t² t' = -1
• C. tt' = -1
• D. None of these

Answer 1

Answer: (A)

t³t' = -1

Answer 2

The equation of the normal to xy = c² at the point 't' is

ty = t³x + c - ct⁴ or t³x - ty + c - ct⁴ = 0 ... (1)

Now, the equation of the line joining 't' and 't' is

$y - \frac{c}{t} = \frac{c|t' - c|t}{ct' - ct}(x - Ct)$

⇒ $\frac{yt - c}{t} = - \frac{1}{tt'}(x - ct)$

⇒ ytt' - ct' = -x + ct

⇒ x + ytt' - c(t + t') = 0

Since (1) and (2) represent the same line,

∴, on comparing the coefficient of x and y, we get

$\frac{1}{t^{3}} = - \frac{tt'}{t} \Rightarrow t^{3}t' = - 1$.