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The equation of the chord joining two points \mathrm{\left(x_1, y_1\right)}  and

\mathrm{\left(x_2, y_2\right)}  on the rectangular hyperbola \mathrm{x y=c^2} is

Option: 1

\mathrm{\frac{x}{x_1+x_2}+\frac{y}{y_1+y_2}=1}


Option: 2

\mathrm{\frac{x}{x_1-x_2}+\frac{y}{y_1-y_2}=1}


Option: 3

\mathrm{\frac{x}{y_1+y_2}+\frac{y}{x_1+x_2}=1}


Option: 4

\mathrm{\frac{x}{y_1-y_2}+\frac{y}{x_1-x_2}=1}


Answers (1)

best_answer

The midpoint of the chord is \mathrm{\left.\left(x_1+x_2\right) / 2,\left(y_1+y_2\right) / 2\right) \text {. }}

The equation of the chord in terms of mid point is 

               \mathrm{T=S_1}

\mathrm{\begin{aligned} & x\left(\frac{y_1+y_2}{2}\right)+y\left(\frac{x_1+x_2}{2}\right)-2 c^2=\left(\frac{x_1+x_2}{2}\right)\left(\frac{y_1+y_2}{2}\right)-2 c^2 \\ & \text { or } x\left(y_1+y_2\right)+y\left(x_1+x_2\right)=\left(x_1+x_2\right)\left(y_1+y_2\right) \\ & \text { or } \frac{x}{x_1+x_2}+\frac{y}{y_1+y_2}=1 \end{aligned}}

Posted by

manish painkra

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