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Particular Solution of a Differential Equation

If the normal...

Question

If the normal at $(c t_{1}, \frac{c}{t_{1}})$ on the hyperbola $x y = c^{2}$ cuts the hyperbola again at $(c t_{2}, \frac{c}{t_{2}})$ , then $t_{2}^{3} t_{2}$ $=$

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Solution

The correct option is C $- 1$
The equation of hyperbola is $x y = c^{2}$ and point $(c t_{1}, \frac{c}{t_{1}})$ lies on it.

Let us find the equation of the normal.

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

hence slope of normal is given by $\frac{x^{2}}{c^{2}}$ and at $(c t_{1}, \frac{c}{t_{1}})$ is $t^{2}$ and its equation is

$y = t_{1}^{2} (x - c t_{1}) + \frac{c}{t_{1}}$

or $x t_{1}^{3} - y t_{1} - c t_{1}^{4} + c = 0$

As this passes through $(c t_{2}, \frac{c}{t_{2}})$

$c t_{2} t_{1}^{3} - \frac{c}{t_{2}} t_{1} - c t_{1}^{4} + c = 0$

or $c t_{1}^{3} (t_{2} - t_{1}) + \frac{c}{t_{2}} (t_{2} - t_{1}) = 0$

as $t_{1} \neq t_{2}, t_{1} - t_{2} \neq 0$ and dividing by it we get

$c t_{2}^{3} = - \frac{c}{t_{2}}$

Or $t_{2}^{3} t_{2} = - 1$

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