For which of the following curves
Question:

For which of the following curves, the line $x+\sqrt{3} y=2 \sqrt{3}$ is the tangent at the point $\left(\frac{3 \sqrt{3}}{2}, \frac{1}{2}\right) ?$

  1. (1) $x^{2}+9 y^{2}=9$

  2. (2) $2 x^{2}-18 y^{2}=9$

  3. (3) $y^{2}=\frac{1}{6 \sqrt{3}} x$

  4. (4) $x^{2}+y^{2}=7$


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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