The equation of the chord joining two points `(x_(1),y_(1))` and `(x_(2),y_(2))` on the rectangular hyperbola `xy=c^(2)`, is

Options

A. `(x)/(x_(1)+x_(2))+(y)/(y_(1)+y_(2))=1`
B. `(x)/(x_(1)-x_(2))+(y)/(y_(1)-y_(2))=1`
C. `(x)/(y_(1)+y_(2))+(y)/(x_(1)+x_(2))=1`
D. `(x)/(y_(1)-y_(2))+(y)/(x_(1)-x_(2))=1`

Answer

A

Explanation

The midpoint of the chord is `((x_(1)+x_(2))//2,(y_(1)+y_(2))//2)`.
The equation of the chord in terms of its midpoint is
`T=S_(1)`
`"or "x((y_(1)+y_(2))/(2))+y((x_(1)+x_(2))/(2))-2c^(2)=((x_(1)+x_(2))/(2))((y_(1)+y_(2))/(2))-2c^(2)`
`"or "x(y_(1)+y_(2))+y(x_(1)+x_(2))=(x_(1)+x_(2))(y_(1)+y_(2))`
`"or "(x)/(x_(1)+x_(2))+(y)/(y_(1)+y_(2))=1`