Find the equation of 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


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