Question

The slopes of common tangents to the hyperbolas

$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ and $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$ are:

• A. ± 2
• B. ± 1
• C. ±$\sqrt{2}$
• D. None of these

Answer 1

Answer: (B)

± 1

Answer 2

Given two hyperbolas are

$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ ... (i)

and $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$ ... (ii)

Equation of tangent on equation (i)having slope m is

y = mx ± $\sqrt{9m^{2} - 16}$ ... (iii)

Eliminating y between equation (ii) and (iii) we get 16

(mx ± $\sqrt{9m^{2} - 16})^{2}$ - 9x² = 144.

⇒ (16m² - 9)x² ± 32m $\sqrt{9m^{2} - 16}$x + (144m² - 400) = 0 ... (iv)

For it to be tangent D = 0.

∴ (32m $\sqrt{9m^{2} - 16})^{2}$ = 4 (16m² - 9) (144m² - 400)

⇒ m² = 1 ⇒ m = ± 1.