Find the equation of the tangents to the ellipse \(3x^2+4y^2\) = 12 which are perpendicular to the line y + 2x = 4.

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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