The equation of the chord joining two points (x1, y1) and (x2, y2) on the rectangular hyperbola x y=c2 is:

Question

The equation of the chord joining two points (x1, y1) and (x2, y2) on the rectangular hyperbola x y=c2 is: (A) (x/x1+x2)+(y/y1+y2)=1 (B) (x/x1-x2)+(y/y1-y2)=1 (C) (x/y1+y2)+(y/x1+x2)=1 (D) (x/y1-y2)+(y/x1-x2)=1

Tardigrade Admin · Accepted Answer

The mid-point of the chord is ((x1+x2/2), (y1+y2/2)). The equation of the chord in terms of its mid-point is T=S1 or x((y1+y2/2))+y((x1+x2/2)) =2((x1+x2/2))((y1+y2/2)) ⇒ x(y1+y2)+y(x1+x2) =(x1+x2)(y1+y2) ⇒ (x/x1+x2)+(y/y1+y2)=1

Q. The equation of the chord joining two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ on the rectangular hyperbola $x y = c^{2}$ is:

$\frac{x}{x _{1} + x _{2}} + \frac{y}{y _{1} + y _{2}} = 1$

$\frac{x}{x _{1} - x _{2}} + \frac{y}{y _{1} - y _{2}} = 1$

$\frac{x}{y _{1} + y _{2}} + \frac{y}{x _{1} + x _{2}} = 1$

$\frac{x}{y _{1} - y _{2}} + \frac{y}{x _{1} - x _{2}} = 1$

Q. The equation of the chord joining two points (x1​,y1​) and (x2​,y2​) on the rectangular hyperbola xy=c2 is:

x1​+x2​x​+y1​+y2​y​=1

x1​−x2​x​+y1​−y2​y​=1

y1​+y2​x​+x1​+x2​y​=1

y1​−y2​x​+x1​−x2​y​=1

Q. The equation of the chord joining two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ on the rectangular hyperbola $x y = c^{2}$ is:

$\frac{x}{x _{1} + x _{2}} + \frac{y}{y _{1} + y _{2}} = 1$

$\frac{x}{x _{1} - x _{2}} + \frac{y}{y _{1} - y _{2}} = 1$

$\frac{x}{y _{1} + y _{2}} + \frac{y}{x _{1} + x _{2}} = 1$

$\frac{x}{y _{1} - y _{2}} + \frac{y}{x _{1} - x _{2}} = 1$