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Parametric Representation: Hyperbola

The normal to...

Question

The normal to the rectangular hyperbola $x y = c^{2}$ at the point ' $t_{1}$ ' meets the curve again at the point ' $t_{2}$ '. Then the value of $t_{1}^{3} t_{2}$ is

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-1

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-c

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Solution

The correct option is B
-1

Normal at $t_{1}$ meets the curve agiain at $t_{2}$ .
So, the normal passes through $(c t_{1}, \frac{c}{t_{1}}) and (c t_{2}, \frac{c}{t_{2}})$
slope of normal = $\frac{\frac{c}{t_{2}} - \frac{c}{t_{1}}}{c t_{2} - c t_{1}} = - \frac{1}{t_{1} t_{2}}$
Also equation of normal at ' $t_{1}$ ' is
$x t_{1}^{3} - y t_{1} - c t_{1}^{4} + c = 0$
$∴$ Slope of normal = $t_{1}^{2} = - \frac{1}{t_{1} t_{2}}$
$∴ t_{1}^{3} t_{2} = - 1$

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Similar questions

x y = - c^{2}

^{'} t_{1}^{'}

^{'} t_{2}^{'}

t_{1}^{3} \cdot t_{2}

^{'} t_{1}^{'}

x y = c^{2}

^{'} t_{2}^{'}

t_{1} t_{2}

x y = 4

t_{1} (= 4)

t_{2}

\frac{1}{| t_{2} |} =

x y = 4

t_{1} (= 4)

t_{2}

\frac{1}{| t_{2} |} =

t_{1}

t_{2}

t_{1}^{3} t_{2}

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