Question

A variable circle whose centre lies on $y^2-36=0$ cuts rectangular hyperbola $xy=16$ at $(4t_i,\frac{4}{t_i}),i=1,2,3,4$ then $\sum _{i=1}^4\frac{1}{t_i}$ can be

• A. $3$
• B. $-3$
• C. $2$
• D. $-2$

Answer 1

Answer: (A), (B)

$-3$

Let circle be $x^2+y^2+2gx+2fy+c=0\cdots \left(1\right)$
It cuts rectangular hyperbola $xy=16\Rightarrow x=\frac{16}{y}\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right),$ we get
$\frac{256}{y^2}+y^2+\frac{32g}{y}+2fy+c=0$
$\Rightarrow y^4+2f{y}^3+c{y}^2+32gy+256=0\cdots \left(3\right)$
Here, we have $4$ solutions for equation $\left(3\right)$
We have, $(x_i,y_i)\equiv (4t_i,\frac{4}{t_i})$
$\therefore y_i=y_1+y_2+y_3+y_4=4(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4})$
$\Rightarrow -2f=4(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4})$
$\frac{-f}{2}=(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4})$
Since, centre lies on $y^2-36=0$
$\therefore f^2=36\Rightarrow f=\pm 6$
Hence, $(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4})=\frac{\pm 6}{2}=\pm 3$