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
$\Rightarrow 4b^4{m}^2{c}^2=4\left(b^2{m}^2-a^2\right)\left(c^2-a^2\right)b^2$
$\Rightarrow c^2=a^2-b^2{m}^2$
$\therefore a^2{m}^2-b^2=a^2-b^2{m}^2$
$\Rightarrow m^2\left(a^2+b^2\right)=a^2+b^2$
$\Rightarrow m=\pm 1$
Hence common tangents are
$y=\pm x\pm \sqrt{a^2-b^2}$