Find Equation of Tangent Line to Hyperbola

Show that the \left( {a\sec \theta ,b\tan \theta } \right) always lies on the hyperbola \frac{{{x^2}}}{{{a^2}}}  - \frac{{{y^2}}}{{{b^2}}} = 1. Find the equation of tangent and normal to the hyperbola \frac{{{x^2}}}{{{a^2}}} -  \frac{{{y^2}}}{{{b^2}}} = 1 at the point \left( {a\sec \theta ,b\tan \theta } \right).
We have standard equation of hyperbola

\frac{{{x^2}}}{{{a^2}}}  - \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,{\text{ -   -  - }}\left( {\text{i}} \right)


Putting x = a\sec  \theta and y = b\tan \theta in equation (i), we have

\begin{gathered} \frac{{{{\left( {a\sec \theta }  \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {b\tan \theta } \right)}^2}}}{{{b^2}}}  = 1 \\ \Rightarrow \frac{{{a^2}{{\sec }^2}\theta  }}{{{a^2}}} - \frac{{{b^2}{{\tan }^2}\theta }}{{{b^2}}} = 1 \\ \Rightarrow {\sec ^2}\theta  - {\tan ^2}\theta  = 1 \\ \end{gathered}


Which is true for all values of \theta , so the point \left( {a\sec \theta ,b\tan \theta } \right) always lies on the hyperbola (i).
Now differentiating equation (i) both sides with respect to x, we have

\begin{gathered} \frac{{2x}}{{{a^2}}} -  \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0 \\ \Rightarrow \frac{{dy}}{{dx}} =  \frac{{{b^2}x}}{{{a^2}y}} \\ \end{gathered}


Let m be the slope of tangent at the given point \left(  {a\sec \theta ,b\tan \theta } \right), then

m =  {\frac{{dy}}{{dx}}_{\left( {a\cos \theta ,b\sin \theta } \right)}} =  \frac{{{b^2}\left( {a\sec \theta } \right)}}{{{a^2}\left( {b\tan \theta }  \right)}} = \frac{{b\sec \theta }}{{a\tan \theta }}


Equation of tangent at the given point \left( {a\sec \theta ,b\tan \theta } \right) is

\begin{gathered} y - b\tan \theta  = \frac{{b\sec \theta }}{{a\tan \theta  }}\left( {x - a\sec \theta } \right) \\ \Rightarrow \frac{{\tan \theta }}{b}\left(  {y - b\tan \theta } \right) = \frac{{\sec \theta }}{a}\left( {x - a\sec \theta  } \right) \\ \Rightarrow \frac{{y\tan \theta }}{b} -  {\tan ^2}\theta  =  - \frac{{x\sec \theta }}{a} - {\sec  ^2}\theta \\ \Rightarrow \frac{x}{a}\sec \theta  - \frac{y}{b}\tan \theta  = {\sec ^2}\theta  - {\tan ^2}\theta \\ \Rightarrow \frac{x}{a}\sec \theta  - \frac{y}{b}\tan \theta  = 1 \\ \end{gathered}


This is the equation of tangent to the given hyperbola at \left( {a\sec \theta ,b\tan \theta } \right).
Slope of the normal at \left(  {a\sec \theta ,b\tan \theta } \right) is  - \frac{1}{m} =   - \frac{{a\tan \theta }}{{b\sec \theta }}
Equation of normal at the point \left( {a\sec \theta ,b\tan \theta } \right) is

\begin{gathered} y - b\tan \theta  =  -  \frac{{a\tan \theta }}{{b\sec \theta }}\left( {x - a\sec \theta } \right) \\ \Rightarrow \frac{b}{{\tan \theta }}\left(  {y - b\tan \theta } \right) = \frac{a}{{\sec \theta }}\left( {x - a\sec \theta  } \right) \\ \Rightarrow \frac{a}{{\sec \theta }}x -  \frac{b}{{\tan \theta }}y = {a^2} - {b^2} \\ \end{gathered}


This is the equation of normal to the given hyperbola at \left( {a\sec \theta ,b\tan \theta } \right).

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