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

Question

The equation of the chord joining two points (x1,y2) and (x2,y2) on the rectangular hyperbola xy=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=S 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 $(x_{1}, y_{2})$ 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​,y2​) 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

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

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

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

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

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

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