Question

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

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

Answer 1

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

Consider two points $(x_1,y_1)$ and $(x_2,y_2)$ lying on the hyperbola $xy=c^2$
We have,
$x_1{y}_1=c^2$ and $x_2{y}_2=c^2$.
Let us consider the equation of the chord between $(x_1,y_1)$ and $(x_2,y_2)$.
$(x-x_1)\left(\frac{y_1-y_2}{x_1-x_2}\right)=y-y_1$
On simplifying we get,
$x\left(\frac{y_1-y_2}{y_1{x}_2-x_1{y}_2}\right)+y\left(\frac{x_2-x_1}{y_1{x}_2-x_1{y}_2}\right)=1$
Substituting $x_1=\frac{c^2}{y_1}$ and $x_2=\frac{c^2}{y_2}$ in the first term and similarly $y_1=\frac{c^2}{x_1}$ and $y_2=\frac{c^2}{x_2}$ in the second term we get,
$x\left(\frac{y_1{y}_2}{c^2(y_1+y_2)}\right)+y\left(\frac{x_1{x}_2}{c^2(x_1+x_2)}\right)=1$
On substituting values of $y_1$ and $y_2$ in terms of $x_1$ and $x_2$ in the first term and $x_1$ and $x_2$ in terms of $y_1$ and $y_2$ in the second term, we get :
$\frac{x}{x_1+x_2}+\frac{y}{y_1+y_2}=1$
Hence, option C is correct.