Question

$PQ$ and $RS$ are two perpendicular chords of the rectangular hyperbola $xy=c^2$. If $C$ is the centre of the rectangular hyperbola, then the product of the slopes of $CP,CQ,CR$ and $CS$ is equal to:

Answer 1

Correct Answer: 1

Explanation:
Given:$PQ$ and $RS$ are two perpendicular chords of the rectangular hyperbola $xy=c^2$ having centre $C$. We have to find the product of the slopes of $CP,CQ,CR$ and $CS$.Let $t_1,t_2,t_3,t_4$ be the parameters of the point $P,Q,R,S$ respectively.Then, the co-ordinates of $P,Q,R,S$ are $(c{t}_1,\frac{c}{t_1}),(c{t}_2,\frac{c}{t_2}),(c{t}_3,\frac{c}{t_3})$ and $(c{t}_4,\frac{c}{t_4})$ respectively, which satisfy the given equation of hyperbola.Now, $PQ\perp RS$$\Rightarrow \frac{\frac{c}{t_2}-\frac{c}{t_1}}{c{t}_2-c{t}_1}\times \frac{\frac{c}{t_4}-\frac{c}{t_3}}{c{t}_4-c{t}_3}=-1$[Product of slopes of perpendicular lines is equal to -1$]$$\Rightarrow \frac{-(t_2-t_1)}{t_1{t}_2(t_2-t_1)}\times \frac{-(t_4-t_3)}{t_3{t}_4(t_4-t_3)}=\frac{1}{t_1{t}_2}\times \frac{1}{t_3{t}_4}=1$$\Rightarrow t_1{t}_2{t}_3{t}_4=1$$$[equation (i)]Now, slope of $CP=\frac{\frac{c}{t_1}-0}{c{t}_1-0}=\frac{1}{t_1^2}$Similarly, we can find slopes of $CQ,CR$ and $CS$ are $\frac{1}{t_2^2},\frac{1}{t_3^2},\frac{1}{t_4^2}$ respectively$\therefore$ Product of the slopes of $CP,CQ,CR$ and $CS$ $=\frac{1}{t_1^2}\times \frac{1}{t_2^2}\times \frac{1}{t_3^2}\times \frac{1}{t_4^2}=\frac{1}{t_1^2{t}_2^2{t}_3^2{t}_4^2}=1[$ using (i) $]$Hence, the correct answer is $1$.