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Equation of the tangent of $$3\mathrm{x}^{2}+4\mathrm{y}^{2}=12$$ parallel to $$\mathrm{x}-2\mathrm{y}+1=0$$ is:


A
x2y+7=0
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B
x2y+4=0
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C
x2y+5=0
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D
x2y+9=0
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Solution

The correct option is B $$x-2y+4=0$$
Equation of line parallel to $$x-2y+1=0$$ is

given by, $$x-2y+c=0$$

Now given ellipse is, $$3x^2+4y^2=12$$

$$\displaystyle\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1 $$

comparing it with standard ellipse $$a^2 = 4, b^2 =3$$

Now using condition of tangency, $$c^2 = a^2m^2+b^2$$

$$\Rightarrow \displaystyle c^2 = 4\times (\dfrac{1}{4})+3=4$$   [$$\because$$ slope of line (1) is $$\dfrac{1}{2}]$$

$$\Rightarrow c = \pm 2$$

Hence required tangent is, $$x-2y\pm 4=0$$

Mathematics

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