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),
$y=mx+c$
Where $c=\pm \sqrt{a^2{m}^2-b^2}$ ...(iii)
But this tangent touches the parabola Eq. (ii), also
$\therefore \frac{(mx+c{)}^2}{a^2}-\frac{x^2}{b^2}=1$
$\Rightarrow b^2\left(m^2{x}^2+c^2+2mcx\right)-a^2{x}^2=a^2{b}^2$
$\Rightarrow \left(b^2{m}^2-a^2\right)x^2+2mcx{b}^2+b^2{c}^2-a^2{b}^2=0$
$\Rightarrow \left(b^2{m}^2-a^2\right)x^2+2mc{b}^{2}x+b^2\left(c^2-a^2\right)=0$
For the tangency, it should have equal roots
${\left(2mc{b}^2\right)}^2=4\left(b^2{m}^2-a^2\right)\cdot b^2\left(c^2-a^2\right)$
$\Rightarrow 4m^2{c}^2{b}^4=4b^2\left(b^2{m}^2{c}^2-b^2{m}^2{a}^2-a^2{c}^2+a^4\right)$
$\Rightarrow m^2{c}^2{b}^2=b^2{m}^2{c}^2-b^2{m}^2{a}^2-a^2{c}^2+a^4$
$\Rightarrow a^2{c}^2=a^4-b^2{m}^2{a}^2$
$\Rightarrow c^2=a^2-b^2{m}^2$
$\Rightarrow a^2{m}^2-b^2=a^2-b^2{m}^2$ [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
$y=\pm x\pm \sqrt{a^2-b^2}$