Equation of Normal to hyperbola : \(x^2\over a^2\) – \(y^2\over b^2\) = 1
(a) Point form :
The equation of normal to the given hyperbola at the point P(\(x_1, y_1\)) is
\(a^2x\over x_1\) + \(b^2y\over y_1\) = \(a^2+b^2\) = \(a^2e^2\)
Example : Find the equation of normal to the hyperbola \(x^2\over 25\) – \(y^2\over 16\) = 1 at (5, 1).
Solution : We have, \(x^2\over 25\) – \(y^2\over 16\) = 1
Compare given equation with \(x^2\over a^2\) – \(y^2\over b^2\) = 1
a = 5 and b = 4
Hence, required equation of normal is \(25x\over 5\) + \(16y\over 1\) = 41
= 5x + 16y = 41
(b) Slope form :
The normal to the given hyperbola whose slope is ‘m’, is
y = mx \(\mp\) \({m(a^2+b^2)}\over \sqrt{a^2 – m^2b^2}\)
Foot of normal are (\(\mp\)\({a^2}\over \sqrt{a^2 – m^2b^2}\), \(\mp\)\({mb^2}\over \sqrt{a^2 – m^2b^2}\)).
Example : Find the normal to the hyperbola \(x^2\over 16\) – \(y^2\over 9\) = 1 whose slope is 1.
Solution : We have, \(x^2\over 16\) – \(y^2\over 9\) = 1
Compare given equation with \(x^2\over a^2\) – \(y^2\over b^2\) = 1
a = 4 and b = 3
Hence, required equation of normal is y = x \(\mp\) \({25}\over \sqrt{7}\)
(c) Parametric form :
The equation of normal to the given hyperbola at its point (asec\(\theta\), btan\(\theta\)), is
\(ax\over sec\theta \) + \(by\over tan\theta \) = \(a^2+b^2\) = \(a^2e^2\)
Example : Line \(xcos\alpha\) + \(ysin\alpha\) = p is a normal to the hyperbola \(x^2\over a^2\) – \(y^2\over b^2\) = 1, if
Solution : We have, \(x^2\over a^2\) – \(y^2\over b^2\) = 1
The normal to hyperbola is \(ax\over sec\theta \) + \(by\over tan\theta \) = \(a^2+b^2\)
comparing it with the given line equation
\(acos\theta\over cos\alpha\) = \(bcot\theta\over sin\alpha\) = \(a^2 + b^2\over p\)
\(\implies\) \(sec\theta\) = \(ap\over cos\alpha(a^2+b^2)\), \(tan\theta\) = \(bp\over sin\alpha(a^2+b^2)\)
Eliminating \(\theta\), we get
\(a^2sec^2\alpha\) – \(b^2cosec^2\alpha\) = \((a^2 + b^2)^2)\over p^2\)
