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  • The equation of the chord joining two points <img alt="\mathrm{\left(x_{1}, y_{1}\right)\: and \: \left(x_{2}, y_{2}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clef

The equation of the chord joining two points \mathrm{\left(x_{1}, y_{1}\right)\: 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{y}{x_{1}+x_{2}}+\frac{x}{y_{1}+y_{2}}=1}


Option: 4

\mathrm{\frac{y}{x_{1}-x_{2}}+\frac{x}{y_{1}-y_{2}}=1}


Answers (1)

best_answer

The midpoint of the chord is \mathrm{\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)}.

The equation of the chord interms of its midpoint is \mathrm{T}=\mathrm{S}^{\prime}

\mathrm{\text { or, } 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) }

\mathrm{\Rightarrow 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 \frac{x}{x_{1}+x_{2}}+\frac{y}{y_{1}+y_{2}}=1}.

Posted by

Suraj Bhandari

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