Question

If the normals at $(x_i,y_i)$, 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)

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

Let $(x_i,y_i)=(\sqrt{2}{t}_i,\frac{\sqrt{2}}{t_i}),i=1,2,3,4$

$\text{equation of normal at}(\sqrt{2}t,\frac{\sqrt{2}}{t})\text{to}xy=2\text{is}$

$y-\frac{\sqrt{2}}{t}=t^2(x-\sqrt{2}t)$

$ty=t^{3}x+\sqrt{2}-\sqrt{2}{t}^4.\text{If it passes through (3,4) then}$

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

$t_1+t_2+t_3+t_4=\frac{3}{\sqrt{2}}$

$x_1+x_2+x_3+x_4=3$

$\sum t_1{t}_2=0\text{and}\sum t_1{t}_2{t}_3=\frac{-4}{\sqrt{2}}$

$t_1{t}_2{t}_3{t}_4=-1\Rightarrow y_1{y}_2{y}_3{y}_4=x_1{x}_2{x}_3{x}_4=-(\sqrt{2}{)}^4=-4$

$y_1+y_2+y_3+y_4=\sqrt{2}[\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4}]$

= $\sqrt{2}\frac{\sum t_1{t}_2{t}_3}{t_1{t}_2{t}_3{t}_4}=4$