Equation of Normal to Hyperbola in all Forms

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

\(a^2sec^2\alpha\) – \(b^2cosec^2\alpha\) = \((a^2 + b^2)^2)\over p^2\)

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