Question

If the normals at $(x_i,y_i),\text{where,}i=1,2,3,4$ on the rectangular hyperbola $xy=c^2$ meet at $(\alpha ,\beta )$. and $x_1^2+x_2^2+x_3^2+x_4^2$ is $a$ and $y_1^2+y_2^2+y_3^2+y_4^2$ is $b$, then $a+b$ is

• A. $\alpha +\beta$
• B. $2(\alpha +\beta )$
• C. ${\alpha }^2+{\beta }^2$
• D. $2({\alpha }^2+{\beta }^2)$

Answer 1

The correct option is C ${\alpha }^2+{\beta }^2$

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 $t_1,t_2,t_3,t_4.$

$\therefore \sum t_1=\frac{\alpha }{c},\sum t_1{t}_2=0\sum \left(t_1{t}_2{t}_3\right)=-\frac{\beta }{c}$
and $\sum \left(t_1{t}_2{t}_3{t}_4\right)=-1.$
$x_1^2+x_2^2+x_3^2+x_4^2=c^2\left[\right(\sum t_1{)}^2-2\sum t_1{t}_2]\therefore a={\alpha }^2$

$c(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4})=y_1+y_2+y_3+y_4=\beta \Rightarrow y_1^2+y_2^2+y_3^2+y_4^2={\beta }^2-2c^2\sum \left(1/t_1{t}_2\right)\therefore b={\beta }^2$

$\therefore a+b={\alpha }^2+{\beta }^2$