Question

The angle between the rectangular hyperbolas (y – mx) (my + x) = a² and (m² – 1) (y² – x²) + 4mxy = b² is-

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (A)

$\frac{\pi}{2}$

Answer 2

For first hyperbola

(y – mx) $\left( m\frac{dy}{dx} + 1 \right)$ + (my + x) $\left( \frac{dy}{dx}–m \right)$ = 0

$\frac{dy}{dx}$ (my + x + my – m²x) + y – mx – m²y – mx = 0

Ž $\frac{dy}{dx}$ = $\frac{- y + m^{2}y + 2mx}{2my + x - m^{2}x}$ = m₁

For second hyperbola

(m² – 1) $\left( 2y\frac{dy}{dx} - 2x \right)$ + 4m $\left( x\frac{dy}{dx} + y \right)$ = 0

$\frac{dy}{dx}$ (2y (m² – 1) + 4 mx) = – 4my + 2x (m² – 1)

Ž $\frac{dy}{dx}$=$\frac{- 2my + m^{2}x - x}{m^{2}y - y + 2mx}$ = m₂ .

Q m₁m₂ = –1 Ž Angle between the hyperbolas = $\frac{\pi}{2}$.

Hence (1) is correct answer.