(2) Substituting P (√ 3/2,3/2) and A (√ 3,0) into parabola respectively, we can get b=√3 and c= 1.
The parabolic equation is: y =-4/3 * x 2+√ 3x+ 1 C point coordinates are (0, 1), which obviously conforms to the equation, so it is on the parabola.
(3)S□MCAP=S△MCP+S△CAP，S△CAP = S△AOC = 1/2 *√3 * 1 =√3/2
The quadrilateral has the largest area, which is equivalent to finding a point on the curve to maximize the area of △MCP.
That is, find a point on the curve to maximize the distance to the edge CP.
It is easy to know from the image that the tangent parallel to CP on the curve has the largest distance from the tangent point to CP on the curve.
Slope k (CP) = (3/2-1)/(√ 3/2-0) = √ 3/3,? The CP equation is y- 1=√3/3*x, that is √3x-3y+3=0.
Derivation of parabola? Y'=-8/3*x+√3 At the tangent point y'=-8/3*x0+√3=√3/3, the solution is x0=√3/4.
Substituting into parabola, y0 =-4/3 * x0 2+√ 3x0+1= 3/2? ∴ The coordinate of point m is (√3/4, 3/2).
The distance from m point to CP line is h = | √ 3 * √ 3/4-3 * 3/2+3 |/√ (3+9) = √ 3/8.
∴s△mcp(max)= 1/2*cp*h= 1/2*oc*h= 1/2* 1*√3/8=√3/ 16
∴s□mcap(max)=s△mcp(max)+s△cap=√3/ 16+√3/2=9√3/ 16？