Question

Normals are drawn from the external point (2,4) to the rectangular hyperbola xy=64. If circles are drawn through the feet of these normals taken three at a time then centre of circle lies on another hyperbola whose centre and recentricity is

• A. (1,2)
  
• B. (2,4)
  
• C. $\sqrt{2}$
• D. $\sqrt{2}+1$

Answer 1

Answer: (A), (C)

The correct options are
A (1,2)

C $\sqrt{2}$

Equation of normal to hyperbola xy=64 at $(8t,\frac{8}{t})$ which passes throught (2,4) is-

$8t^4-2t^3+4t-8=0$
this equation has 4 roots $t_1,t_2,t_3\text{and}{t}_4$
$t_1{t}_2{t}_3{t}_4$ = -1

Let equation of circle be $x^2+y^2-2gx-2fy+e=0$

If $(8t,\frac{8}{t})$ lies on this cirlce

$t_1{t}_2{t}_3{t}_4^1=1\dots \left(2\right)$

From (1) & (2) $\Rightarrow t_4=-t_4^1$

$\&t_1{t}_2{t}_3{t}_4(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4})=-\frac{1}{2}\Rightarrow \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4}=\frac{1}{2}\dots \left(3\right)$

$t_1{t}_2{t}_3{t}_4(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4})=-\frac{16f}{64}=\frac{f}{4}\Rightarrow \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4}=\frac{f}{2}\dots \left(4\right)$

Equation (3)-(4)

$\frac{1}{f_4}-\frac{1}{f_4^1}=\frac{1}{2}-\frac{f}{4}\Rightarrow \frac{1}{f_4}+\frac{1}{f_4}=\frac{1}{2}-\frac{f}{4}\Rightarrow \frac{2}{f_4}=\left(\frac{2-f}{4}\right)\dots \left(5\right)$

$t_1+t_2+t_3+t_4=\frac{1}{4}$

subtract $\frac{t_1+t_2+t_3+t_4=\frac{g}{4}}{t_4-t_4^1=\frac{1-g}{4}}$

$2t_4=\frac{1-g}{4}\dots \left(6\right)$

Multiplying equation (5) & (6)

$4=(\frac{2-f}{4}\times \frac{1-g}{4})$
Centre $\to (g,f)=(h,k)$

64=(h-1)(k-2)

Since it is Rectangular hyperbola so eccentity is $\sqrt{2}$