Answer:
[tex]x+\sqrt{3}y=8[/tex]
Step-by-step explanation:
Given equation of curve,
[tex]y^2+x^2=16[/tex]
[tex]\implies y^2=16-x^2[/tex]
Differentiating with respect to x,
[tex]2y\frac{dy}{dx}=-2x[/tex]
[tex]\implies \frac{dy}{dx}=-\frac{x}{y}[/tex]
Since, the tangent line of the curve contains the point (2, 2√3),
Thus, the slope of the tangent line,
[tex]m=\left [ \frac{dy}{dx} \right ]_{(2, 2\sqrt{3})}=-\frac{1}{\sqrt{3}}[/tex]
Hence, the equation of tangent line would be,
[tex]y-2\sqrt{3}=-\frac{1}{\sqrt{3}}(x-2)[/tex]
[tex]\sqrt{3}y-6=-x+2[/tex]
[tex]\implies x+\sqrt{3}y=8[/tex]
