Q65 / 120 · VITEEE 2010 · Mathematics
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$y=\pm x \pm \sqrt{b^{2}-a^{2}}$
- Correct answerB$y=\pm x \pm \sqrt{a^{2}-b^{2}}$
- C$y=\pm x \pm \sqrt{a^{2}+b^{2}}$
- D$y=\pm x \pm \sqrt{a^{2}-b^{2}}$
Solution
We have the two hyperbolas as
$$
\begin{array}{l}
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1...(i) \\
\text { and } \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1...(ii)
\end{array}
$$
Any tangent to the hyperbola eq(i) $y=m x+c$
where $c=\pm \sqrt{a^{2} m^{2}-b^{2}} \ldots$ (iii)
But this tangent touches the parabola eq. (ii) also
$$
\begin{array}{l}
\therefore \frac{(m x+c)^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1 \\
\Rightarrow b^{2}\left(m^{2} x^{2}+c^{2}+2 m c x\right)-a^{2} x^{2}=a^{2} b^{2} \\
\Rightarrow\left(b^{2} m^{2}-a^{2}\right) x^{2}+2 m c b^{2} x+b^{2}\left(c^{2}-a^{2}\right)=0
\end{array}
$$
For the tangency, it should have equal roots
$$
\begin{aligned}
&\left(2 m c b^{2}\right)^{2}=4\left(b^{2} m^{2}-a^{2}\right) \cdot b^{2}\left(c^{2}-a^{2}\right) \\
\Rightarrow & 4 m^{2} c^{2} b^{4}=4 b^{2}\left(b^{2} m^{2} c^{2}-b^{2} m^{2} a^{2}-a^{2} c^{2}+a^{4}\right) \\
\Rightarrow & c^{2}=a^{2}-b^{2} m^{2} \\
\Rightarrow & a^{2} m^{2}-b^{2}=a^{2}-b^{2} m^{2}[\text { using Eq. (iii) }] \\
\Rightarrow &\left(a^{2}+b^{2}\right) m^{2}=a^{2}+b^{2} \\
\Rightarrow & m^{2}=1 \Rightarrow m=\pm 1
\end{aligned}
$$
Hence, the equation of common tangent are
$$
y=\pm x \pm \sqrt{a^{2}-b^{2}}
$$
Hence, the correct answer is option: 'B'.