Question

The equation of the common tangents to the two hyperbolas $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} =1 $ and $ \frac{y^{2}}{a^{2} } - \frac{x^{2}}{b^{2}} = 1 $, are

• A. $y = \pm x \pm \sqrt{b^{2} -a^{2}} $
• B. $y = \pm x \pm \sqrt{a^{2} - b^{2}} $
• C. $y = \pm x \pm \sqrt{a^{2} + b^{2}} $
• D. $y = \pm x \pm \sqrt{b^{2} - a^{2}} $

Answer 1

Answer: (B)

We have the hyperbolas
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ...(i)
and $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ ...(ii)
Any tangent to the hyperbola Eq. (i),
$y=m x +c$
Where $c=\pm \sqrt{a^{2} m^{2}-b^{2}}$ ...(iii)
But this tangent touches the parabola Eq. (ii), also
$\therefore \frac{(m x+c)^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$
$\Rightarrow b^{2}\left(m^{2} x^{2}+c^{2}+2 m c x\right)-a^{2} x^{2}=a^{2} b^{2}$
$\Rightarrow \left(b^{2} m^{2}-a^{2}\right) x^{2}+2 m c x b^{2}+b^{2} c^{2}-a^{2} b^{2}=0$
$\Rightarrow \left(b^{2} m^{2}-a^{2}\right) x^{2}+2 m c b^{2} x +b^{2}\left(c^{2}-a^{2}\right)=0$
For the tangency, it should have equal roots
$\left(2 m c b^{2}\right)^{2}=4\left(b^{2} m^{2}-a^{2}\right) \cdot b^{2}\left(c^{2}-a^{2}\right)$
$\Rightarrow 4 m^{2} c^{2} b^{4}=4 b^{2}\left(b^{2} m^{2} c^{2}-b^{2} m^{2} a^{2}-a^{2} c^{2}+a^{4}\right)$
$\Rightarrow m^{2} c^{2} b^{2}=b^{2} m^{2} c^{2}-b^{2} m^{2} a^{2}-a^{2} c^{2}+a^{4}$
$\Rightarrow a^{2} c^{2}=a^{4}-b^{2} m^{2} a^{2}$
$\Rightarrow c^{2}=a^{2}-b^{2} m^{2}$
$\Rightarrow a^{2} m^{2}-b^{2}=a^{2}-b^{2} m^{2}$ [using Eq.
$\Rightarrow \left(a^{2}+b^{2}\right) m^{2}=a^{2}+b^{2}$
$\Rightarrow m^{2}=1$
$\Rightarrow m=\pm 1$
Hence, the equation of common tangent are
$y=\pm x \pm \sqrt{a^{2}-b^{2}}$