Examples on Tangents To Hyperbolas Set 1
Example - 12
Tangents are drawn to the hyperbola \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) at the points \({\theta _1}\) and \({\theta _2}\) Find their point of intersection.
Solution : The equations of the two tangents are
\[\frac{x}{a}\sec {\theta _1} - \frac{y}{b}\tan {\theta _1} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\]
\[\frac{x}{a}\sec {\theta _2} - \frac{y}{b}\tan {\theta _2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\]
(1) ×\(\tan {\theta _2}\) – (2) ×\(\tan {\theta _1}\) gives
\[\frac{x}{a}(\sec {\theta _1}\tan {\theta _2} - \sec {\theta _2}\tan {\theta _1}) = \tan {\theta _2} - \tan {\theta _1}\]
Simplifying, this yields
\[x = \frac{{a\cos \left( {\begin{align}\frac{{{\theta _1} - {\theta _2}}}{2}\end{align}} \right)}}{{\cos \left({\begin{align} \frac{{{\theta _1} + {\theta _2}}}{2}\end{align}} \right)}}\qquad\qquad\qquad \ldots \,\,\left( 3 \right)\]
(1) \( \times \sec {\theta _2} - \) (2) \( \times \sec {\theta _1}\) gives
\[\frac{y}{b}(\sec {\theta _1}\tan {\theta _2} - \sec {\theta _2}\tan {\theta _1}) = \sec {\theta _2} - \sec {\theta _1}\]
Simplifying, this gives
\[y = \frac{{b\sin \left( {\begin{align}\frac{{{\theta _1} + {\theta _2}}}{2}\end{align}} \right)}}{{\cos \left( {\begin{align}\frac{{{\theta _1} + {\theta _2}}}{2}\end{align}}\right)}}\qquad\qquad\qquad \ldots \left( 4 \right)\]
From (3) and (4), the point of intersection P of the two tangents is
\[P \equiv \left( {\frac{{a\cos \left( {\begin{align}\frac{{{\theta _1} - {\theta _2}}}{2}\end{align}} \right)}}{{\cos \left( {\begin{align}\frac{{{\theta _1} + {\theta _2}}}{2}\end{align}} \right)}},\,\,\frac{{b\sin \left( {\begin{align}\frac{{{\theta _1} + {\theta _2}}}{2}\end{align}} \right)}}{{\cos \left( {\begin{align}\frac{{{\theta _1} + {\theta _2}}}{2}\end{align}} \right)}}} \right)\]
Example - 13
Find the condition that must be satisfied if the line \(px + qy + r = 0\) is to be a tangent to the hyperbola \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.\end{align}\)
Solution : Any tangent to this hyperbola is of the form
\[mx - y + \sqrt {{a^2}{m^2} - {b^2}} = 0\]
If the line \(px + qy + r = 0\) is also a tangent, we must have, for some real m,
\[\begin{align} & \;\;\;\;\;\;\;\;\;\frac{p}{m} = \frac{q}{{ - 1}} = \frac{r}{{\sqrt {{a^2}{m^2} - {b^2}} }} \\ &\Rightarrow \quad m = \frac{{ - p}}{q}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{from the first two relations}}} \right) \\&\Rightarrow \quad {a^2}{m^2} - {b^2} = \frac{{{r^2}}}{{{q^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{from the last two relations}}} \right) \\ & \Rightarrow \quad \frac{{{a^2}{p^2}}}{{{q^2}}} - {b^2} = \frac{{{r^2}}}{{{q^2}}} \hfill \\ &\Rightarrow \quad {a^2}{p^2} - {b^2}{q^2} = {r^2}. \end{align} \]
This is the requisite condition that must be satisfied.