The equation of the common tangents to the&nb | vedclass

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Question

Any tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is

$y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}$

or $y=m x+c$

$	ext { where

Vedclass · Accepted Answer

$y = &plusmn; x &plusmn; \sqrt {a^2 - b^2}$

Vedclass · Answer

Any tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is

$y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}$

or $y=m x+c$

$	ext { 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 $\frac{(m x+c)^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ has equal roots or $\mathrm{x}^{2}\left(\mathrm{b}^{2} \mathrm{m}^{2}-\mathrm{a}^{2}ight)+2 \mathrm{b}^{2} \mathrm{mc} \mathrm{x}+\left(\mathrm{c}^{2}-\mathrm{a}^{2}ight) \mathrm{b}^{2}=0$ is an quadratic eqaution have equal roots 

then $4 b^{4} m^{2} c^{2}=4\left(b^{2} m^{2}-a^{2}ight)\left(c^{2}-a^{2}ight) b^{2}$

or $c^{2}=a^{2}-b^{2} m^{2}$

$\quad a^{2} m^{2}-b^{2}=a^{2}-b^{2} m^{2} \quad$ put $\left(c^{2}=a^{2} m^{2}-b^{2}ight)$

$m^{2}\left(a^{2}+b^{2}ight)=a^{2}+b^{2} \quad \Rightarrow \quad m=\pm 1$

The equation of the common tangents to the two hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are-

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