Question

If the tangent and normal to a rectangular hyperbola cut off intercepts $a_{1}$ and $a_{2}$ on one axis and $b_{1}$ and $b_{2}$ on the other axis, then

• A. $a_{1}b_{1} + a_{2}b_{2} = 0$
• B. $a_{1}b_{2} + b_{2}a_{1} = 0$
• C. $a_{1}a_{2} + b_{1}b_{2} = 0$
• D. None of these

Answer 1

Answer: (C)

$a_{1}a_{2} + b_{1}b_{2} = 0$

Answer 2

Let the hyperbola be $xy = c^{2}$. Tangent at any point t is $x + yt^{2} - 2ct = 0$

Putting $y = 0$ and then $x = 0$ intercepts on the axes are $a_{1} = 2ct$ and $b_{1} = \frac{2c}{t}$

Normal is $xt^{3} - yt - ct^{4} + c = 0$.

Intercepts as above are $a_{2} = \frac{c(t^{4} - 1)}{t^{3}}$, $b^{2} = \frac{- c(t^{4} - 1)}{t}$

$\therefore$ $a_{1}a_{2} + b_{1}b_{2} = 2ct \times \frac{c(t^{4} - 1)}{t^{3}} + \frac{2c}{t} \times \frac{- c(t^{4} - 1)}{t}$ =

$\frac{2c^{2}}{t^{2}}(t^{4} - 1) - \frac{2c^{2}}{t^{2}}(t^{4} - 1) = 0$; $\therefore$ $a_{1}a_{2} + b_{1}b_{2} = 0$