On a point (ct,c/t) on hyperbola xy=c^2, a normal is drawn which intersects the hyperbola at (ct',c/(t')). Prove that t^3t'=-1?

Jul 28, 2017

Explanation:

The equation of hyperbola is $x y = {c}^{2}$ and point $\left(c t , \frac{c}{t}\right)$ lies on it.

Let us find the equation of the normal.

Equation of hyprbola can be written as $y = {c}^{2} / x$ and therefore slope of tangent is given by first derivative i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = - {c}^{2} / {x}^{2}$

hence slope of normal is given by ${x}^{2} / {c}^{2}$ and at $\left(c t , \frac{c}{t}\right)$ is ${t}^{2}$ and its equation is

$y = {t}^{2} \left(x - c t\right) + \frac{c}{t}$

or $x {t}^{3} - y t - c {t}^{4} + c = 0$

As this passes through $\left(c t ' , \frac{c}{t '}\right)$

$c t ' {t}^{3} - \frac{c}{t '} t - c {t}^{4} + c = 0$

or $c {t}^{3} \left(t ' - t\right) + \frac{c}{t '} \left(t ' - t\right) = 0$

as $t \ne t '$, $t - t ' \ne 0$ and dividing by it we get

$c {t}^{3} = - \frac{c}{t '}$

or ${t}^{3} t ' = - 1$