# Condition for Line Tangent to a Hyperbola

The condition for a line $y = mx + c$ to be the tangent to the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ is that $c = \pm \sqrt {{a^2}{m^2} – {b^2}}$ and the tangent to the hyperbola is $y = mx \pm \sqrt {{a^2}{m^2} – {b^2}}$.

Consider the equation of a line is represented by

$y = mx + c\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Consider that the standard equation of a hyperbola with vertex at origin $\left( {0,0} \right)$ can be written as

$\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

To find the point of intersection of straight line (i) and the given hyperbola (ii), using the method of solving simultaneous equations we solve equation (i) and equation (ii). Putting the value of $y$ from equation (i) in equation (ii), we have

$\begin{gathered} \frac{{{x^2}}}{{{a^2}}} – \frac{{{{\left( {mx + c} \right)}^2}}}{{{b^2}}} = 1 \\ \Rightarrow \frac{{{b^2}{x^2} – {a^2}{{\left( {mx + c} \right)}^2}}}{{{a^2}{b^2}}} = 1 \\ \Rightarrow {b^2}{x^2} – {a^2}{\left( {mx + c} \right)^2} = {a^2}{b^2} \\ \Rightarrow {b^2}{x^2} – {a^2}\left( {{m^2}{x^2} + 2mcx + {c^2}} \right) = {a^2}{b^2} \\ \Rightarrow {b^2}{x^2} – {a^2}{m^2}{x^2} – 2{a^2}mcx – {a^2}{c^2} – {a^2}{b^2} = 0 \\ \Rightarrow \left( {{a^2}{m^2} – {b^2}} \right){x^2} + 2{a^2}mcx + {a^2}\left( {{b^2} + {c^2}} \right) = 0\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right) \\ \end{gathered}$ If equation (iii) has equal roots, then the line equation (i) will intersect the hyperbola (ii) at one point only and thus is the tangent to the hyperbola.

For equal roots, we have

$\begin{gathered} {\text{Discriminant = 0}} \\ \Rightarrow {\left( {2{a^2}mc} \right)^2} – 4\left( {{a^2}{m^2} – {b^2}} \right){a^2}\left( {{b^2} + {c^2}} \right) = 0 \\ \Rightarrow 4{a^4}{m^2}{c^2} – 4{a^2}\left( {{a^2}{m^2} – {b^2}} \right)\left( {{b^2} + {c^2}} \right) = 0 \\ \Rightarrow {a^2}{m^2}{c^2} – \left( {{a^2}{m^2} – {b^2}} \right)\left( {{b^2} + {c^2}} \right) = 0 \\ \Rightarrow – {a^2}{m^2} + {b^2} + {c^2} = 0 \\ \Rightarrow {c^2} = {a^2}{m^2} – {b^2} \\ \Rightarrow c = \pm \sqrt {{a^2}{m^2} – {b^2}} \\ \end{gathered}$

Putting these values of $c$ in the equation of straight line (i), we have
$y = mx \pm \sqrt {{a^2}{m^2} – {b^2}}$

These are the tangents to the hyperbola as shown in the given diagram.