Question

If the circle $x^2+y^2=a^2$ intersects the hyperbola $xy=c^2$ in four points $P(x_1,y_1),Q(x_2,y_2)R(x_3,y_3),S(x_4,y_4)$, then which of the following need not hold?

• A. $x_1+x_2+x_3+x_4=0$
• B. $x_1{x}_2{x}_3{x}_4=y_1{y}_2{y}_3{y}_4=c^4$
• C. $y_1+y_2+y_3+y_4=0$
• D. $x_1+y_2+y_3+y_4=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A

$x_1+x_2+x_3+x_4=0$

B

$x_1{x}_2{x}_3{x}_4=y_1{y}_2{y}_3{y}_4=c^4$

C

$y_1+y_2+y_3+y_4=0$

Any point on $xy=c^2\text{is}(ct,\frac{c}{t})$ which lies on the circle $x^2+y^2=a^2$
$\Rightarrow c^2{t}^2+\frac{c^2}{t^2}=a^2\Rightarrow c^2{t}^4-a^2{t}^2+c^2=0$
$t_1,t_2,t_3,t_4$ are the roots of this $4^{\text{th}}$ order equation. Sum of roots taken one at a time = $-\frac{b}{a}$
$\Rightarrow t_1+t_2+t_3+t_4=0$
Multiplying the above equation by $c$ we get,
$x_1+x_2+x_3+x_4=0\text{where}{x}_i=c{t}_i,i=1,2,3,4.$
Product of all roots = $\frac{e}{a}$
$\Rightarrow t_1{t}_2{t}_3{t}_4=1\Rightarrow x_1{x}_2{x}_3{x}_4=y_1{y}_2{y}_3{y}_4=c^4$
Product of roots taken three at a time = $-\frac{d}{a}$
$\Rightarrow t_1{t}_2{t}_3+t_1{t}_2{t}_4+t_1{t}_3{t}_4+t_2{t}_3{t}_4=0\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4}=0\Rightarrow y_1+y_2+y_3+y_4=0$