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

Option: 1

\mathrm{a_1 b_1+a_2 b_2=0}


Option: 2

\mathrm{a_1 b_2+b_2 a_1=0}


Option: 3

\mathrm{a_1 a_2+b_1 b_2=0}


Option: 4

None of these


Answers (1)

best_answer

Let the hyperbola be \mathrm{xy=c^{2}}. Tangent at any point \mathrm{t} is \mathrm{x+y t^2-2 c t=0}
Putting \mathrm{y=0} and then \mathrm{x=0}  intercepts on the axes are \mathrm{a_1=2 c t \text { and } b_1=\frac{2 c}{t}}
Normal is \mathrm{x t^3-y t-c t^4+c=0}
Intercepts as above are \mathrm{a_2=\frac{c\left(t^4-1\right)}{t^3} \quad b^2=\frac{-c\left(t^4-1\right)}{t}}
\mathrm{\therefore \quad a_1 a_2+b_1 b_2=2 c t \times \frac{c\left(t^4-1\right)}{t^3}+\frac{2 c}{t} \times \frac{-c\left(t^4-1\right)}{t}=\frac{2 c^2}{t^2}\left(t^4-1\right)-\frac{2 c^2}{t^2}\left(t^4-1\right)=0}
\mathrm{\therefore a_1 a_2+b_1 b_2=0}

Posted by

Pankaj Sanodiya

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