Question

The equation of the chord joining two points $(x_1,y_1)$ and $(x_2,y_2)$ on rectangular hyperbola $xy=c^2$ is

• A. $\frac{x}{(x_1+x_2)}+\frac{y}{y_1+y_2}=1$
• B. $\frac{x}{x_1-x_2}+\frac{y}{y_1-y_2}=1$
• C. $\frac{x}{y_1+y_2}+\frac{y}{x_1+x_2}=1$
• D. $\frac{x}{y_1-y_2}+\frac{y}{x_1-x_2}=1$

Answer 1

The correct option is A $\frac{x}{(x_1+x_2)}+\frac{y}{y_1+y_2}=1$

Well, you can find out equation by using basic rules of Geometry

Because two points $(x_1,y_1)$ and $(x_2,y_2)$ are given then,

Slope of chord is $(y_2-y_1)(x_2,x_1)$

Now, equation of chord is $(y-y_1)=(y_2-y_1)(x_2-x_1)(x-x_1)$

But we have to use term c in equation of chord.

so. let's start.

can we write $(x_1,y_1)=(x_1.c^2/x_1)[\because x_1{y}_1=c^2\Rightarrow y_1=c^2/x_1]$

Similarly, $(x_2-y_2)=(x_2,c^2/x_2)$

Now, slope of chord is $c^2(x_1-x_2)/x_1{x}_2(x_2-x_1)=-c^2/x_1{x}_2$

So, equation of chord is given by

$(y-c^2/x_1)+c^2/x_1{x}_2(x-x_1)=0$

$y-c^2/x_1+c^2/x_1{x}_2,x-c^2/x_2=0$

$x_1{x}_{2}y+c^{2}x=(x_1+x_2)c^2$

$x_1{x}_{2}y+c^{2}x=c^2(x_1+x_2)$

$\frac{x_1{x}_{2}y}{c^2(x_1+x_2)}+\frac{x}{x_1+x_2}=1$

$\frac{y}{y_1+y_2}+\frac{x}{x_1+x_2}=1$ [as $\frac{c^2}{x_1}=y_1]$