Question

If the normals at $(x_i,y_i),$ where, $i=1,2,3,4$ on the rectangular hyperbola $xy=c^2$ meet at $(\alpha ,\beta )$, then

• A. $x_1+x_2+x_3+x_4=\beta$
• B. $y_1+y_2+y_3+y_4=\beta$
• C. $x_1+x_2+x_3+x_4=\alpha$
• D. $y_1+y_2+y_3+y_4=\alpha$

Answer 1

Answer: (B), (C)

The correct options are
B $y_1+y_2+y_3+y_4=\beta$
C $x_1+x_2+x_3+x_4=\alpha$

The equation of the normal to the hyperbola $xy=c^2$ at $(ct,\frac{c}{t})$ is
$x{t}^3-yt-c{t}^4+c=0$
which is passing through $(\alpha ,\beta )$
Thus, $c{t}^4-\alpha t^3+\beta t-c=0.$
Let its four roots are $t_1,t_2,t_3,t_4.$

$\therefore t_1+t_2+t_3+t_4=\frac{\alpha }{c},\sum \left(t_1{t}_2{t}_3\right)=-\frac{\beta }{c}$
and $\left(t_1{t}_2{t}_3{t}_4\right)=-1.$

Now
$c{t}_1+c{t}_2+c{t}_3+c{t}_4=\alpha$
$\Rightarrow x_1+x_2+x_3+x_4=\alpha$
and,
$c(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4})=\beta \Rightarrow y_1+y_2+y_3+y_4=\beta$