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.