For which of the following curves, the line

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. $x^{2}+y^{2}=7$

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

  3. $2 x^{2}-18 y^{2}=9$

  4. $x^{2}+9 y^{2}=9$


Correct Option: , 4

Solution:

$\mathrm{m}=-\frac{1}{\sqrt{3}}, \mathrm{c}=2$

(1) $\mathrm{c}=\mathrm{a} \sqrt{1+\mathrm{m}^{2}}$

$\mathrm{c}=\sqrt{7} \frac{2}{\sqrt{3}}$ (incorrect)

(2) $\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}=\frac{\frac{1}{24 \sqrt{3}}}{\frac{-1}{\sqrt{3}}}=-\frac{1}{24}$ (incorrect)

(3) $c=\sqrt{a^{2} m^{2}-b^{2}}$

$\mathrm{c}=\sqrt{\frac{9}{2} \cdot \frac{1}{3}-\frac{1}{2}}=1 \quad$ (incorrect)

(4) $\mathrm{c}=\sqrt{\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}}$

$\mathrm{c}=\sqrt{9 \cdot \frac{1}{3}+1}=2 \quad$ (correct)

 

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