Question

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

• 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 (a^2-b^2) $
• D. $ y = \pm x \pm \sqrt{a^2+b^2}$

Answer 1

Answer: (B)

Any tangent to the hyperbola
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $y= mx \pm \sqrt{ a^{2}m^{2} -b^{2}} $
or $y= mx+c$ where $c = \pm \sqrt{ a^{2}m^{2} -b^{2}}$
This will touch the hyperbola $ \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$
if the equation $\frac{\left(mx+c\right)^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 $ has equal roots
or $x^{2}\left(b^{2}m^{2} -a^{2}\right) + 2b^{2}mcx $
$ + \left(c^{2}-a^{2}\right)b^{2} = 0$ has equal roots
$\Rightarrow 4b^{4}m^{2}c^{2} = 4\left(b^{2}m^{2} -a^{2}\right)\left(c^{2}-a^{2}\right)b^{2}$
$\Rightarrow c^{2} = a^{2}-b^{2}m^{2} $
$ \therefore a^{2}m^{2} -b^{2} = a^{2}-b^{2}m^{2}$
$ \Rightarrow m^{2}\left(a^{2}+b^{2}\right) = a^{2}+b^{2} $
$\Rightarrow m =\pm1 $
Hence common tangents are
$ y= \pm x \pm \sqrt{ a^{2}-b^{2}}$