Question

The variable chords of the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 (b > a)

whose equation is

x \cos α + y \sin α = p

subtends a right angle at the centre. Prove that it always touches a circle.

Accepted Answer

For the point of intersection of $x \cos α + y \sin α = p$ and $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{x^{2}}{a^{2}} - \frac{{(\frac{p - x \cos α}{\sin α})}^{2}}{b^{2}} = 1$

$\Rightarrow \frac{x^{2}}{a^{2}} - \frac{{(p - x \cos α)}^{2}}{b^{2} \sin^{2} α} = 1$

$\Rightarrow b^{2} x^{2} \sin^{2} α - a^{2} {(p - x \cos α)}^{2} = a^{2} b^{2} \sin^{2} α$

$\Rightarrow b^{2} x^{2} \sin^{2} α - a^{2} (p^{2} + x^{2} \cos^{2} α - 2 x p \cos α) = a^{2} b^{2} \sin^{2} α$

$\Rightarrow (b^{2} \sin^{2} α - a^{2} \cos^{2} α) x^{2} + (2 a^{2} p \cos α) x - a^{2} p^{2} - a^{2} b^{2} \sin^{2} α = 0$

Let, $A (x_{1}, \frac{p - x_{1} \cos α}{\sin α}) & B (x_{2}, \frac{p - x_{2} \cos α}{\sin α})$ are the point of intersection of $x \cos α + y \sin α = p & \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

$∴ x_{1} + x_{2} = - \frac{2 a^{2} p \cos α}{b^{2} \sin^{2} α - a^{2} \cos^{2} α} & x_{1} x_{2} = \frac{- a^{2} p^{2} - a^{2} b^{2} \sin^{2} α}{b^{2} \sin^{2} α - a^{2} \cos^{2} α}$ .

Let, $C$ is the centre of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and $m_{1} & m_{2}$ are the slopes of the lines $C A & C B$ .

$\Rightarrow m_{1} = \frac{\frac{p - x_{1} \cos α}{\sin α} - 0}{x_{1} - 0} & m_{2} = \frac{\frac{p - x_{2} \cos α}{\sin α} - 0}{x_{2} - 0}$

$\Rightarrow m_{1} m_{2} = \frac{1}{\sin^{2} α} \times \frac{p^{2} - p (x_{1} + x_{2}) \cos α + x_{1} x_{2} \cos^{2} α}{x_{1} x_{2}}$

$\Rightarrow m_{1} m_{2} = \frac{1}{\sin^{2} α} \times \frac{p^{2} + p (\frac{2 a^{2} p \cos α}{b^{2} \sin^{2} α - a^{2} \cos^{2} α}) \cos α + (\frac{- a^{2} p^{2} - a^{2} b^{2} \sin^{2} α}{b^{2} \sin^{2} α - a^{2} \cos^{2} α}) \cos^{2} α}{\frac{- a^{2} p^{2} - a^{2} b^{2} \sin^{2} α}{b^{2} \sin^{2} α - a^{2} \cos^{2} α}}$

$\Rightarrow m_{1} m_{2} = - \frac{1}{\sin^{2} α} \times \frac{p^{2} (b^{2} \sin^{2} α - a^{2} \cos^{2} α) + p (2 a^{2} p \cos α) \cos α - (a^{2} p^{2} + a^{2} b^{2} \sin^{2} α) \cos^{2} α}{a^{2} p^{2} + a^{2} b^{2} \sin^{2} α}$

$\Rightarrow m_{1} m_{2} = - \frac{p^{2} b^{2} - a^{2} b^{2} \cos^{2} α}{a^{2} p^{2} + a^{2} b^{2} \sin^{2} α}$

$\Rightarrow m_{1} m_{2} = - \frac{p^{2} b^{2} - a^{2} b^{2} (1 - \sin^{2} α)}{a^{2} p^{2} + a^{2} b^{2} \sin^{2} α}$

$\Rightarrow m_{1} m_{2} = - 1$

Hence, the lines $C A & C B$ are perpendicular to each other.

We also know that $x \cos α + y \sin α = p$ is a tangent to the circle $x^{2} + y^{2} = p^{2}$ at $(p \cos α, p \sin α)$ .

Embibe Experts Solutions for Chapter: Hyperbola, Exercise 4: EXERCISE-4

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Hyperbola, Exercise 4: EXERCISE-4

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