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  • If the tangent and normal to a rectangular hyperbola cutoff intercepts a and b on one axis and p and q on another axis, then,  Option: 1 ap

If the tangent and normal to a rectangular hyperbola cutoff intercepts a and b on one axis and p and q on another axis, then,  

Option: 1

ap - bq = 0


Option: 2

ap + bq = 0


Option: 3

ab - pq = 0


Option: 4

ab + pq = 0


Answers (1)

best_answer

 

 

Properties of rectangular Hyperbola -

Properties of rectangular Hyperbola:

 

\\\mathrm{(i)\;\;\;\text{The parametric equation of the rectangular hyperbola}\;xy=c^2}\\\mathrm{\;\;\;\;\;\;\;\;\;are\;\;x=ct\;\;and\;\;y=\frac{c}{t}.}\\\\\mathrm{(ii)\;\;\;\text{The equation of the tangent to the rectangular hyperbola}\;xy=c^2}\\\mathrm{\;\;\;\;\;\;\;\;\;at\;(x_1,y_1)\;\;is\;\;x y_{1}+x_{1} y=c^{2}.}\\\\\mathrm{(iii)\;\;\;\text { The equation of the tangent at }\left(c t, \frac{c}{t}\right) \text { to the hyperbola }\;xy=c^2}\\\mathrm{\;\;\;\;\;\;\;\;\;\text { is } \frac{x}{t}+y t=2 c.}\\\\\mathrm{(iv)\;\;\;\text { The equation of the normal at }\left(x_{1}, y_{1}\right) \text { to the hyperbola }\;xy=c^2}\\\mathrm{\;\;\;\;\;\;\;\;\;\text { is } x x_{1}-y y_{1}=x_{1}^{2}-y_{1}^{2}.}\\\\\mathrm{(v)\;\;\;\text { The equation of the normal at }t \text { to the hyperbola }\;xy=c^2}\\\mathrm{\;\;\;\;\;\;\;\;\;\text { is } x t^{3}-y t-c t^{4}+c=0.}

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Let rectangular hyperbola is xy = c2

equation of the tangent at point 't' is

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}\frac{x}{t}+yt=2c\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\frac{x}{2ct}+\frac{y}{\frac{2c}{t}}=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots ( i)

Equation of normal at point 't' is

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}xt^3-yt-ct^4+c=0\\\Rightarrow \;\;\;\;\;\;\;\;\;xt^3-yt=ct^4-c\\\Rightarrow \;\;\;\;\;\;\;\;\;\frac{x}{\left ( \frac{ct^4-c}{t^3} \right )}+\frac{y}{\left ( \frac{-ct^4+c}{t} \right )}=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)

frm (i) and (ii)

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}a=2ct,\;\;\;b=\frac{ct^4-c}{t^3}\\\text{and}\;\;\;\;\;\;\;\;\;\;\;\;\;\;p=\frac{2c}{t},\;\;\;\;q=\frac{-ct^4+c}{t}\\\\\therefore \;\;\;\;\;ab+pq\\\mathrm{\;\;\;\;\;\;}\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{2ct(ct^4-c)}{t^3}+\frac{2c}{t}\cdot\frac{(-ct^4+c)}{t}=0

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