The equations of the common tangents to the t | vedclass

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(D) Let the two hyperbolas be $H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $H_2: \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.Any tangent to $H_1$ is gi

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(D) Let the two hyperbolas be $H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $H_2: \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.Any tangent to $H_1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.For this line to be a tangent to $H_2$,we rewrite $H_2$ as $\frac{x^2}{(-b^2)} - \frac{y^2}{(-a^2)} = 1$. The condition for tangency $y = mx + c$ to $\frac{x^2}{A} - \frac{y^2}{B} = 1$ is $c^2 = Am^2 - B$.Here $A = -b^2$ and $B = -a^2$,so $c^2 = (-b^2)m^2 - (-a^2) = a^2 - b^2m^2$.Equating the two expressions for $c^2$: $a^2m^2 - b^2 = a^2 - b^2m^2$.$(a^2 + b^2)m^2 = a^2 + b^2$,which gives $m^2 = 1$,so $m = \pm 1$.Substituting $m^2 = 1$ into $c^2 = a^2m^2 - b^2$,we get $c^2 = a^2 - b^2$,so $c = \pm \sqrt{a^2 - b^2}$.Thus,the common tangents are $y = \pm x \pm \sqrt{a^2 - b^2}$.All given options are included in this set.

The equations 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:

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