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

Since the normal to the given hyperbola whose slope is ‘m’, is  y = mx \(\mp\) \({m(a^2+b^2)}\over \sqrt{a^2 – m^2b^2}\)

Hence, required equation of normal is y = x \(\mp\) \({25}\over \sqrt{7}\).

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