Question

The equations of the common tangents to the two hyperbolas

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a<b)$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1(a>b)$ is

• A. $y=\pm x\pm \sqrt{a^2+b^2}$
• B. $y=\pm x\pm \sqrt{a^2-b^2}$
• C. $y=\pm x\pm \sqrt{a^3+b^3}$
• D. $y=\pm x\pm \sqrt{a^3-b^3}$

Answer 1

The correct option is B $y=\pm x\pm \sqrt{a^2-b^2}$

Let $y=mx+c$ be the common tangent of the given hyperbolas.
Since, $y=mx+c$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\Rightarrow c^2=a^2{m}^2-b^2\dots \left(1\right)$
Since, $y=mx+c$ is a tangent to the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\Rightarrow c^2=a^2-b^2{m}^2$
$\Rightarrow a^2{m}^2-b^2=a^2-b^2{m}^2\left[\text{using}\right(1\left)\right]$

$\Rightarrow a^2{m}^2+b^2{m}^2=a^2+b^2$

$\Rightarrow m^2(a^2+b^2)=(a^2+b^2)$

$\Rightarrow m^2=1$ and $m=\pm 1$
Now, from (1)

$c^2=a^2{m}^2-b^2$

$\Rightarrow c^2=a^2(1{)}^2-b^2$

$\Rightarrow c=\pm \sqrt{a^2-b^2}$
Hence tangents will be,

$y=mx+c$
$\Rightarrow y=\pm x\pm \sqrt{a^2-b^2}$