Tangent Paralleled?

The best way to do is to use the fact that the equations of tangent, with slope $m$ , to standard hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ are $y=mx\pm \sqrt{a^2m^2-b^2}$

Now the distance between lines $y=mx+c_1$ and $y=mx+c_2$ is $\left |\dfrac{c_1-c_2}{\sqrt{1+m^2}}\right |$

Substituting $a^2=9,b^2=49$ , equations of tangents come out to be $y=mx\pm\sqrt{9m^2-49}$

Calculating distance and equating to $2$ ,

$2=\dfrac{2\sqrt{9m^2-49}}{\sqrt{1+m^2}}\\\Rightarrow \sqrt{1+m^2}=\sqrt{9m^2-49}\\\Rightarrow 1+m^2=9m^2-49\Rightarrow m^2=\dfrac{25}{4}\\\Rightarrow |m|=\dfrac{5}{2}\Rightarrow 2|m|=5$

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Hint for proving the equation of tangents

Let the equation of line be $y=mx+c$ . Solve it with hyperbola to get a quadratic in $x$ or $y$ . Since it is a tangent, there should be one solution to this quadratic. Make the determinant $0$ to get a quadratic in $c$ . Calculate $c$ as $f(m,a,b)$ . Substitute $c$ back in original equation. $\ddot\smile$

By symmetry each tangent must be 1 unit from the origin. Giving y intercept to be $\frac{1}{\cos\theta} = \sec \theta = \sqrt{1 + m^2}$ . Equate this with the y intercept from the equation of tangent: $y = mx \pm \sqrt{a^2m^2-b^2}$

Thus, $1 + m^2 = 9m^2-49$ yielding $m = \frac{5}{2}$