Question

The slopes of the common tangents to the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ and $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$ are

• A. – 2, 2
• B. – 1, 1
• C. 1, 2
• D. 2, 1

Answer 1

Answer: (B)

– 1, 1

Answer 2

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

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

Any tangent to (i) having slope m is $y = mx \pm \sqrt{9m^{2} - 16}$ .....(iii)

Putting in (ii), we get, $16\lbrack mx \pm \sqrt{9m^{2} - 16}\rbrack^{2} - 9x^{2} = 144$⇒

$(16m^{2} - 9)x^{2} \pm 32m(\sqrt{9m^{2} - 16})x + 144m^{2} - 256 - 144 = 0$⇒ $(16m^{2} - 9)x^{2} \pm 32m(\sqrt{9m^{2} - 16})x + (144m^{2} - 400) = 0$.....(iv)

If (iii) is a tangent to (ii), then the roots of (iv) are real and equal.

$\therefore$Discriminant = 0; $32 \times 32m^{2}(9m^{2} - 16)$ =

$4(16m^{2} - 9)(144m^{2} - 400)$ = $64(16m^{2} - 9)(9m^{2} - 25)$

⇒$16m^{2}(9m^{2} - 16) = (16m^{2} - 9)(9m^{2} - 25)$

⇒ $144m^{4} - 256m^{2} = 144m^{4} - 481m^{2} + 225$

⇒ $225m^{2} = 225$⇒ $m^{2} = 1$ ⇒ $m = \pm 1$