Answer
\[y=\frac{3}{2}x+\frac{1}{2}\text{ and }y=-3x-8\text{ }\]
Work Step by Step
\[\begin{align}
& y=x\sqrt{5-{{x}^{2}}} \\
& \text{Differentiate} \\
& \frac{dy}{dx}=\frac{d}{dx}\left[ x\sqrt{5-{{x}^{2}}} \right] \\
& \frac{dy}{dx}=x\frac{d}{dx}\left[ \sqrt{5-{{x}^{2}}} \right]+\sqrt{5-{{x}^{2}}}\frac{d}{dx}\left[ x \right] \\
& \frac{dy}{dx}=x\left( \frac{-2x}{2\sqrt{5-{{x}^{2}}}} \right)+\sqrt{5-{{x}^{2}}}\left( 1 \right) \\
& \frac{dy}{dx}=\sqrt{5-{{x}^{2}}}-\frac{{{x}^{2}}}{\sqrt{5-{{x}^{2}}}} \\
& \text{Calculate the derivative at the point }\left( 1,2 \right)\text{ and }\left( -2,-2 \right) \\
& {{\left. \frac{dy}{dx} \right|}_{x=1}}=\sqrt{5-{{\left( 1 \right)}^{2}}}-\frac{{{\left( 1 \right)}^{2}}}{\sqrt{5-{{\left( 1 \right)}^{2}}}}=\frac{3}{2} \\
& {{\left. \frac{dy}{dx} \right|}_{x=-2}}=\sqrt{5-{{\left( -2 \right)}^{2}}}-\frac{{{\left( -2 \right)}^{2}}}{\sqrt{5-{{\left( -2 \right)}^{2}}}}=-3 \\
& \text{Find the equation of the tangent line at the point }\left( 1,2 \right)\text{ } \\
& \text{and }\left( -2,-2 \right) \\
& y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\
& y-2=\frac{3}{2}\left( x-1 \right) \\
& y-2=\frac{3}{2}x-\frac{3}{2} \\
& \text{ }y=\frac{3}{2}x+\frac{1}{2} \\
& and \\
& y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\
& y+2=-3\left( x+2 \right) \\
& y+2=-3x-6 \\
& \text{ }y=-3x-8 \\
& \text{Graph} \\
\end{align}\]
