The equation of a common tangent to the conic | vedclass

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(C) Let the equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $y = mx \pm \sqrt{a^2m^2 - b^2}$.For the hyperbola $\f

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

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(C) Let the equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $y = mx \pm \sqrt{a^2m^2 - b^2}$.For the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,which can be written as $\frac{x^2}{(-b^2)} - \frac{y^2}{(-a^2)} = 1$,the condition for tangency $y = mx + c$ is $c^2 = a^2m^2 - b^2$ for the first and $c^2 = -a^2 + b^2m^2$ for the second.Equating the squared intercepts: $a^2m^2 - b^2 = b^2m^2 - a^2$.$(a^2 - b^2)m^2 = b^2 - a^2$.Since $a^2 - b^2 
eq 0$,we have $m^2 = -1$,which is not possible for real tangents.However,considering the symmetry $y = x$ or $y = -x$,for the curves to have a common tangent,we set $m = \pm 1$.Substituting $m = 1$ into $c^2 = a^2(1)^2 - b^2$,we get $c = \sqrt{a^2 - b^2}$.Thus,the equation of the common tangent is $y = x \pm \sqrt{a^2 - b^2}$,or $x - y = \pm \sqrt{a^2 - b^2}$.

The equation of a common tangent to the conics $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ is

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