Solution :
Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4
\(\therefore\) mx – 2 = -1 \(\implies\) m = \(1\over 2\)
Since \(3x^2+4y^2\) = 12 or \(x^2\over 4\) + \(y^2\over 3\) = 1
Comparing this with \(x^2\over a^2\) + \(y^2\over b^2\) = 1
\(\therefore\) \(a^2\) = 4 and \(b^2\) = 3
So the equation of the tangent are y = \(1\over 2\)x \(\pm\) \(\sqrt{4\times {1\over 4} + 3}\)
\(\implies\) y = \(1\over 2\)x \(\pm\) 2 or x – 2y \(\pm\) 4 = 0
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