Question:
For which of the following curves, the line $x+\sqrt{3} y=2 \sqrt{3}$ is the tangent at the point $\left(\frac{3 \sqrt{3}}{2}, \frac{1}{2}\right) ?$
Correct Option: 1,
Solution:
Tangent to $x^{2}+9 y^{2}=9$ at point $\left(\frac{3 \sqrt{3}}{2}, \frac{1}{2}\right)$ is $\times\left|\frac{3 \sqrt{3}}{2}\right|+9 y\left(\frac{1}{2}\right)=9$
$3 \sqrt{3} \mathrm{x}+9 \mathrm{y}=18 \Rightarrow \mathrm{x}+\sqrt{3} \mathrm{y}=2 \sqrt{3}$
$\Rightarrow$ option (1) is true
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