# Normals To Hyperbolas

Like tangents, a normal to any hyperbola can also be specified in one of many possible forms. As before, we will use the hyperbola \begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}
for this discussion.

NORMAL AT (x1, y1): In the previous section, we obtained the derivative of the hyperbola at P(x1, y1) as

${\left. {\frac{{dy}}{{dx}}} \right|_{P({x_1},{y_1})}} = \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}$

Therefore, the slope of the normal at $$P({x_1},\,\,{y_1})$$ is

${m_N} = \frac{{ - {a^2}{y_1}}}{{{b^2}{x_1}}}$

The equation of the normal at P is thus

$y - {y_1} = \frac{{ - {a^2}{y_1}}}{{{b^2}{x_1}}}(x - {x_1})$

which upon rearrangement gives

$\boxed{\frac{{{a^2}x}}{{{x_1}}} + \frac{{{b^2}y}}{{{y_1}}} = {a^2} + {b^2}}$

NORMAL AT $$P{\mathbf{(}}a\,{\mathbf{sec}}\,{\mathbf{\theta }},\,\,b\,{\mathbf{tan}}\,{\mathbf{\theta )}}$$ : Again, if the point P has been specified in parametric form, the equation of the normalcan be obtained by the substitutions $${x_1} \to a\sec \theta$$ and $${y_1} \to b\tan \theta$$ in the equation obtained above:

\begin{align}&\;\;\;\;\;\;\qquad\frac{{{a^2}x}}{{a\sec \theta }} + \frac{{{b^2}y}}{{b\tan \theta }} = {a^2} + {b^2} \\\\&\Rightarrow \qquad\boxed{ax\sin \theta + by = ({a^2} + {b^2})\tan \theta }\\\end{align}

NORMAL OF SLOPE m : Let $$P({x_1},\,\,{y_1})$$ be the point at which a line of slope m is normal to the hyperbola.

Thus,

\begin{align}&\qquad\;\;\;m = \frac{{ - {a^2}{y_1}}}{{{b^2}{x_1}}} \\\\&\Rightarrow \quad {y_1} = \frac{{ - {b^2}m}}{{{a^2}}}{x_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right) \\ \end{align}

Since $$P({x_1},{y_1})$$  must satisfy the equation of the hyperbola, we have

\begin{align}&\qquad\;\;\frac{{x_1^2}}{{{a^2}}} - \frac{{y_1^2}}{{{b^2}}} = 1 \\ \\ &\Rightarrow \quad \frac{{x_1^2}}{{{a^2}}} - \frac{{{b^2}{m^2}}}{{{a^4}}}x_1^2 = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {Using{\text{ }}\left( 1 \right)} \right) \\ \\ &\Rightarrow \quad {x_1} = \pm \frac{{{a^2}}}{{\sqrt {{a^2} - {b^2}{m^2}} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.\,..\left( 2 \right) \\\\&\Rightarrow \quad {y_1} = \mp \frac{{{b^2}m}}{{\sqrt {{a^2} - {b^2}{m^2}} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {From{\text{ }}\left( 1 \right){\text{ }}and{\text{ }}\left( 2 \right)} \right) \end{align}

Thus, the equation of the normal at P is

\begin{align}&\qquad\qquad\frac{{{a^2}x}}{{{x_1}}} + \frac{{{b^2}y}}{{{y_1}}} = {a^2} + {b^2} \\ &\Rightarrow\qquad \pm \left( {\sqrt {{a^2} - {b^2}{m^2}} x - \frac{{\sqrt {{a^2} - {b^2}{m^2}} }}{m}y} \right) = {a^2} + {b^2} \\ &\Rightarrow\qquad \boxed{y = mx \mp \frac{{m({a^2} + {b^2})}}{{\sqrt {{a^2} - {b^2}{m^2}} }}} \\ \end{align}

This is (are) the equation (s) of the normal (s) to the hyperbola with slope m. We can say from this relation that a line will be a normal to the hyperbola \begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align} if

\begin{align}{c^2} = \frac{{{m^2}({a^2} + {b^2})}}{{{a^2} - {b^2}{m^2}}}\end{align}

Example - 16

Show that four normals( real or imaginary) can be drawn from any point P to a given hyperbola, say \begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.\end{align}  In addition, show that the sum of the eccentric angles of the feet of these four normals is an odd multiple of  $$\pi .$$

Solution : We encountered a very similar question in the unit on Ellipse, and we’ll follow exactly the same approach here. Any normal to the hyperbola  \begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align} has the form

$(a\cos \theta )x + (b\cot \theta )y = {a^2} + {b^2}$

If this normal passes through the given point P whose coordinates we can assume to be (h, k), we have

$(a\cos \theta )h + (b\cot \theta )k = {a^2} + {b^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)$

We have to show that this equation will in general be satisfied by four values of $$\theta .$$ For that, we use the substitution

$\cos \theta \to \frac{{1 - {t^2}}}{{1 + {t^2}}};\,\,\cot \theta \to \frac{{1 - {t^2}}}{{2t}};\,\,t = \tan \frac{\theta }{2}$

This when used in (1) gives

\begin{align}&\qquad\;\;\;\;\;\;\;ah\left( {\frac{{1 - {t^2}}}{{1 + {t^2}}}} \right) + bk\left( {\frac{{1 - {t^2}}}{{2t}}} \right) = {a^2} + {b^2} \\\\ &\Rightarrow\qquad bk{t^4} + 2(ah + {a^2} + {b^2}){t^3} + 2({a^2} + {b^2} - ah)t - bk = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\ldots \left( 2 \right) \\ \end{align}

This biquadratic equation in t will yield four roots ( say $${t_1},\,\,{t_2},\,\,{t_3},\,\,{t_4})$$ and hence four normals are possible ( of course, all the four roots may not be real in which case less than four normals are physically possible).We now come to the second part of the question. From (2), we have

\begin{align}&{S_1} = \sum {{t_i} = \frac{{ - 2(ah + {a^2} + {b^2})}}{{bk}}}\\ &{S_2} = \sum {{t_i}{t_j} = 0} \\\\ &{S_3} = \sum {{t_i}{t_j}{t_k} = \frac{{2(ah - {a^2} - {b^2})}}{{bk}}} \\\\&{S_4} = \sum {{t_1}{t_2}{t_3}{t_4} = - 1} \\ \end{align}

Thus, we have

$\tan \left( {\frac{{{\theta _1}}}{2} + \frac{{{\theta _2}}}{2} + \frac{{{\theta _3}}}{2} + \frac{{{\theta _4}}}{2}} \right) = \frac{{{S_1} - {S_3}}}{{1 - {S_2} + {S_4}}} = \frac{{({\text{finite)}}}}{{{\text{(zero)}}}}\\\\\qquad\;\;=\infty$

\begin{align} & \Rightarrow\qquad \frac{{{\theta _1} + {\theta _2} + {\theta _3} + {\theta _4}}}{2} = n\pi + \frac{\pi }{2}\\& \Rightarrow \qquad\;\; {\theta _1} + {\theta _2} + {\theta _3} + {\theta _4} = (2n + 1)\pi \end{align}

This final relation proves the assertion stated in the question.