Answer
\[y=\frac{\sqrt{1-2x^2}}{2}\]
Work Step by Step
Slope of the tangent to a curve at ($x,y$) is given by $\frac{dy}{dx}$
\[\frac{dy}{dx}=\frac{-x}{4y}\]
Seperating variables
\[4y dy=-x dx\]
Integrating
\[4\int y dy=-\int x dx+C\]
Where C is constant of integration
\[\frac{4y^2}{2}=\frac{-x^2}{2}+C\]
$2y^2+x^2=C$ ________(1)
Curve passes through (0,$\frac{1}{2}$)
$\frac{2}{4}=\frac{1}{2}=C$
From (1)
\[2y^2=\frac{1}{2}-x^2=\frac{1-2x^2}{2}\]
\[\Rightarrow y=\frac{\sqrt{1-2x^2}}{2}\]
Hence equation of curve is $y=\frac{\sqrt{1-2x^2}}{2}$
