Equation of Tangent to Hyperbola \(x^2\over a^2\) – \(y^2\over b^2\) = 1
(a) Point form :
The equation of tangent to the given hyperbola at its point (\(x_1, y_1\)) is
\(x{x_1}\over a^2\) – \(y{y_1}\over b^2\) = 1
Example : Find the equation of tangent to the hyperbola \(16x^2\) – \(9y^2\) = 144 at (5, 16/3).
Solution : We have, \(16x^2\) – \(9y^2\) = 144
\(\implies\) \(x^2\over 9\) – \(y^2\over 16\) = 1
Compare given equation with \(x^2\over a^2\) – \(y^2\over b^2\) = 1
a = 3 and b = 16
Hence, required equation of tangent is \(5x\over 9\) – \(16/3\over 16\) = 1
= \(5x\over 9\) – \(y\over 3\) = 1
(b) Slope form :
The equation of tangent to the given hyperbola whose slope is ‘m’, is
y = mx \(\pm\) \(\sqrt{a^2m^2 – b^2}\)
The Point of contact are (\({\mp} a^2m\over \sqrt{a^2m^2 – b^2}\), \({\mp} b^2\over \sqrt{a^2m^2 – b^2}\))
Note that there are two parallel tangents having the same slope m.
Example : Find the tangent to the hyperbola \(x^2\over 25\) – \(y^2\over 16\) = 1 whose slope is 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 y = x \(\pm\) \(\sqrt{9}\)
\(\implies\) y = x \(\pm\) 3
(c) Parametric form :
The equation of tangent to the given hyperbola at the point (asec\(\theta\), btan\(\theta\)), is
\(xsec\theta\over a\) – \(ytan\theta\over b\) = 1
Example : Find the tangent to the hyperbola \(x^2 – 4y^2\) = 36 which is perpendicular to the line x – y + 4 = 0
Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line x – y = 0
\(\therefore\) m\(\times\)1 = -1 \(\implies\) m = -1
Since \(x^2-4y^2\) = 36 or \(x^2\over 36\) – \(y^2\over 9\) = 1
Comparing this with \(x^2\over a^2\) – \(y^2\over b^2\) = 1
\(\therefore\) \(a^2\) = 36 and \(b^2\) = 9
So the equation of the tangent are y = -1x \(\pm\) \(\sqrt{36\times {-1}^2 – 9}\)
\(\implies\) y = x \(\pm\) \(\sqrt{27}\) \(\implies\) x + y \(\pm\) 3\(\sqrt{3}\) = 0
