# 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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