Question

If the normal’s at $(x_i,y_i)$ where $i=1,2,3,4$ to the rectangular hyperbola $xy=2$ meet at the point (3, 4), then

• A. $x_1+x_2+x_3+x_4=3$
• B. $y_1+y_2+y_3+y_4=4$
• C. $x_1{x}_2{x}_3{x}_4=-4$
• D. $y_1{y}_2{y}_3{y}_4=-4$

Answer 1

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

The correct options are
A

$x_1+x_2+x_3+x_4=3$

B

$y_1+y_2+y_3+y_4=4$

C

$x_1{x}_2{x}_3{x}_4=-4$

D

$y_1{y}_2{y}_3{y}_4=-4$

Any point on xy = 2 is $P(\sqrt{2}t,\frac{\sqrt{2}}{t})$

$xy=2$

$\Rightarrow x\cdot \frac{dy}{dx}+y\cdot 1=0$

$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}=-\frac{1}{t^2}$

Normal at P is $(y-\frac{\sqrt{2}}{t})=t^2(x-\sqrt{2}t)$

Normal passes through (3,4)

$\therefore \sqrt{2}{t}^4-3t^3+4t-\sqrt{2}=0$

$\sum t_1=\frac{3}{\sqrt{2}}$

$\sum t_1{t}_2=0$

$\sum t_1{t}_2{t}_3=-2\sqrt{2}$

$\sum t_1{t}_2{t}_3{t}_4=-1$