1. The problem asks for the equation of a circle with center at the origin and radius $s$. 2. The general equation for a circle with center at the origin $(0,0)$ and radius $r$ is: $$x^2 + y^2 = r^2$$ 3. Substituting the radius $s$ into the equation, we get: $$x^2 + y^2 = s^2$$ This is the equation of the circle. 4. Next, we need to reduce the equation $2x^2 - x + 4y = 3$ to standard form. 5. Start with the given equation: $$2x^2 - x + 4y = 3$$ 6. Isolate the $y$ term: $$4y = 3 - 2x^2 + x$$ 7. Divide both sides by 4: $$y = \frac{3}{4} - \frac{2}{4}x^2 + \frac{1}{4}x = \frac{3}{4} - \frac{1}{2}x^2 + \frac{1}{4}x$$ 8. Rearrange the terms: $$y = -\frac{1}{2}x^2 + \frac{1}{4}x + \frac{3}{4}$$ 9. This is a quadratic in $x$ with $y$ expressed as a function of $x$, representing a parabola. 10. To write in standard form, complete the square for the $x$ terms. 11. Factor out $-\frac{1}{2}$ from the $x^2$ and $x$ terms: $$y = -\frac{1}{2}\left(x^2 - \frac{1}{2}x\right) + \frac{3}{4}$$ 12. Complete the square inside the parentheses: Take half of $-\frac{1}{2}$ which is $-\frac{1}{4}$, square it gives $\left(-\frac{1}{4}\right)^2 = \frac{1}{16}$. Add and subtract $\frac{1}{16}$ inside the parentheses: $$y = -\frac{1}{2}\left( x^2 - \frac{1}{2}x + \frac{1}{16} - \frac{1}{16} \right) + \frac{3}{4}$$ 13. Group to form a perfect square and simplify: $$y = -\frac{1}{2}\left( x - \frac{1}{4} \right)^2 + \frac{1}{32} + \frac{3}{4}$$ 14. Combine constants: $$\frac{1}{32} + \frac{3}{4} = \frac{1}{32} + \frac{24}{32} = \frac{25}{32}$$ 15. Final standard form: $$y = -\frac{1}{2}\left( x - \frac{1}{4} \right)^2 + \frac{25}{32}$$ This is the standard form of the parabola.