Question

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

Answer 1

The correct option is B $y=\pm x\pm (a^2-b^2)$

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 $c=\pm \sqrt{a^2{m}^2-b^2}$ ...(1)
This will touch the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ if the equation $\frac{{(mx+c)}^2}{a^2}-\frac{x^2}{b^2}=1$ has equal roots.
or $x^2(b^2{m}^2-a^2)+2b^{2}mcx+(c^2-a^2)b^2=0$ has equal roots
$\Rightarrow 4b^4{m}^2{c}^2=4(b^2{m}^2-a^2)(c^2-a^2)b^2\Rightarrow c^2=a^2-b^2{m}^2$
$\therefore a^2{m}^2-b^2=a^2-b^2{m}^2$ ...(Using (1))
$\Rightarrow m^2(a^2+b^2)=a^2+b^2\Rightarrow m=\pm 1$
Hence, the equation of common tangent are $y=\pm x\pm \sqrt{a^2-b^2}$