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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