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  • The equation of the chord joining two points <img alt="\mathrm{\left(x_1, y_1\right) \text { and }\left(x_2, y_2\right)}" src="https://entrancecorner.oncodecogs.com/gif.l

The equation of the chord joining two points \mathrm{\left(x_1, y_1\right) \text { and }\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 mid-point of the chord is \mathrm{\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)}

The equation of the chord in terms of its mid-point is

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

Posted by

Sanket Gandhi

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