HyperbolaHardBloom L3

Question

If the normal to the rectangular hyperbola $xy = c^{2}$ at the point ' $t$ ' then $t^{3}t_{1}$ has the value equal to :

Options

A.1
B.-1
C.0
D.none of these

Solution

Sol. (b)Slope of normal at $A$

$$\begin{matrix} & \ = - \frac{1}{\frac{dy}{dx}} = \frac{x^{2}}{c^{2}} = t^{2} \\ & \text{~}\text{of chord}\text{~}AB \end{matrix}$$

$$\begin{matrix} & \ = \frac{c\left( \frac{1}{t} - \frac{1}{t_{1}} \right)}{c\left( t - t_{1} \right)} = \frac{- 1}{tt_{1}} \\ \Rightarrow \ t^{2} & \ = - \frac{1}{tt_{1}} \end{matrix}$$

$$\Rightarrow \ t^{3}t_{1} = - 1$$

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