Normal to rectangular hyperbola xy = c^2 at point 't₁' meets curve again at point 't?

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10-2. Parabola, Ellipse, Hyperbola

normal

The normal to the rectangular hyperbola $xy = c^2$ at the point $'t_1'$ meets the curve again at the point $'t_2'$ . Then the value of $t_{1}^{3} t_{2}$ is

A

$1$

B

$c$

C

$-c$

D

$-1$

Solution

Normal at $t_{1}$ meets the curve again at $t_{2}.$

So normal passes through point $\left(c t_{1}, c / t_{1}\right)$ and $\left(c t_{2}, c / t_{2}\right).$

Slope of normal

$=\frac{c / t_{2}-c / t_{1}}{c t_{2}-c t_{1}}=-\frac{1}{t_{1} t_{2}}$

Slope of normal from equation of normal at $t_{1}=t_{1}^{2}$

So $t_{1}^{2}=-\frac{1}{t_{1} t_{2}} \Rightarrow t_{1}^{3} t_{2}=-1$

Standard 11

Mathematics

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