$P Q$ and $R S$ are two perpendicular chords of the rectangular hyperbola $x y = c^{2}$ . If $C$ is the centre of the rectangular hyperbola, then the product of the slopes of $C P, C Q, C R$ and $C S$ is equal to:

Question

If the line $ ax + 2y = 1 $, where $ a \in \mathbb{R} $, does not meet the hyperbola $ x^2 - 9y^2 = 9 $, then a possible value of $ a $ is:

Answer 1

To solve the problem, we need to find the product of the slopes of the lines

C\!P

,

C\!Q

,

C\!R

, and

C\!S

for the rectangular hyperbola defined by

xy=c^2

, where

PQ

and

RS

are perpendicular chords.

The center

C

of a hyperbola

xy=c^2

is the origin (0,0). Let's represent the points on the hyperbola as

P(x_1,\frac{c^2}{x_1})

,

Q(x_2,\frac{c^2}{x_2})

,

R(x_3,\frac{c^2}{x_3})

, and

S(x_4,\frac{c^2}{x_4})

.

The slopes of lines from the origin to these points are:

m_{CP}=\frac{\frac{c^2}{x_1}-0}{x_1-0}=\frac{c^2}{x_1^2},\ m_{CQ}=\frac{\frac{c^2}{x_2}-0}{x_2-0}=\frac{c^2}{x_2^2},\ m_{CR}=\frac{\frac{c^2}{x_3}-0}{x_3-0}=\frac{c^2}{x_3^2},\ m_{CS}=\frac{\frac{c^2}{x_4}-0}{x_4-0}=\frac{c^2}{x_4^2}.

Now, find the product of these slopes:

m_{CP}\cdot m_{CQ}\cdot m_{CR}\cdot m_{CS}=\frac{c^2}{x_1^2}\cdot\frac{c^2}{x_2^2}\cdot\frac{c^2}{x_3^2}\cdot\frac{c^2}{x_4^2} = \frac{c^8}{x_1^2x_2^2x_3^2x_4^2}.

Since

PQ

and

RS

are perpendicular chords of a rectangular hyperbola, by geometric properties, the product of their slopes is -1, thus:

\left(\frac{c^2}{x_1^2}\cdot\frac{c^2}{x_2^2}\right) \left(\frac{c^2}{x_3^2}\cdot\frac{c^2}{x_4^2}\right) = -1.

Therefore, calculating the product of the derived individual slopes for chords gives us:

(m_{CP}\cdot m_{CQ})\cdot(m_{CR}\cdot m_{CS}) = -1 \cdot (-1) = 1.

So, the required product satisfies the range 1, 1. Therefore, the final answer is: 1.

Correct Answer: 1