In the accompanying figure , $y=f(x)$ is the graph of a one-to-one continuous function $f$ . At each point $P$ on the graph of $y=2x^2$ , assume that the areas $OAP$ and $OBP$ are equal . Here $PA,PB$ are the horizontal and vertical segments . Determine the function $f$. 


[asy]
Label f;  
xaxis(0,60,blue); 
yaxis(0,60,blue); 
real f(real x) 
{ 
return (x^2)/60; 
} 
draw(graph(f,0,53),red);
label("$y=x^2$",(30,15),E);
real f(real x) 
{ 
return (x^2)/25; 
} 
draw(graph(f,0,38),red);
label("$y=2x^2$",(37,37^2/25),E);
real f(real x) 
{ 
return (x^2)/10; 
} 
draw(graph(f,0,25),red);
label("$y=f(x)$",(24,576/10),W);
label("$O(0,0)$",(0,0),S);
dot((20,400/25));
dot((20,400/60));
label("$P$",(20,400/25),E);
label("$B$",(20,400/60),SE);
dot(((4000/25)^(0.5),400/25));
label("$A$",((4000/25)^(0.5),400/25),W);
draw((20,400/25)..((4000/25)^(0.5),400/25));
draw((20,400/25)..(20,400/60));
[/asy]
1. **Define Points and Areas:**
   - Let the points \(A\), \(P\), and \(B\) be \((t, f(t))\), \((\sqrt{\frac{f(t)}{2}}, f(t))\), and \((\sqrt{\frac{f(t)}{2}}, \frac{f(t)}{2})\) respectively.
   - Denote the areas by \([ \ ]\).

2. **Express Areas \(OAP\) and \(OBP\):**
   - The area \([OAP]\) can be expressed as:
     \[
     [OAP] = \int_0^t f(x) \, dx + f(t) \left( \sqrt{\frac{f(t)}{2}} - t \right) - \int_0^{\sqrt{\frac{f(t)}{2}}} 2x^2 \, dx
     \]
   - The area \([OBP]\) can be expressed as:
     \[
     [OBP] = \int_0^{\sqrt{\frac{f(t)}{2}}} 2x^2 \, dx - \int_0^{\sqrt{\frac{f(t)}{2}}} x^2 \, dx
     \]

3. **Set Areas Equal:**
   - Since \([OAP] = [OBP]\), we equate the two expressions:
     \[
     \int_0^t f(x) \, dx + f(t) \left( \sqrt{\frac{f(t)}{2}} - t \right) - \int_0^{\sqrt{\frac{f(t)}{2}}} 2x^2 \, dx = \int_0^{\sqrt{\frac{f(t)}{2}}} 2x^2 \, dx - \int_0^{\sqrt{\frac{f(t)}{2}}} x^2 \, dx
     \]

4. **Simplify the Integrals:**
   - Calculate the integrals:
     \[
     \int_0^{\sqrt{\frac{f(t)}{2}}} 2x^2 \, dx = \left[ \frac{2x^3}{3} \right]_0^{\sqrt{\frac{f(t)}{2}}} = \frac{2}{3} \left( \sqrt{\frac{f(t)}{2}} \right)^3 = \frac{f(t)^{3/2}}{3 \sqrt{2}}
     \]
     \[
     \int_0^{\sqrt{\frac{f(t)}{2}}} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{\sqrt{\frac{f(t)}{2}}} = \frac{1}{3} \left( \sqrt{\frac{f(t)}{2}} \right)^3 = \frac{f(t)^{3/2}}{6 \sqrt{2}}
     \]

5. **Substitute and Simplify:**
   - Substitute the integrals back into the equation:
     \[
     \int_0^t f(x) \, dx + f(t) \left( \sqrt{\frac{f(t)}{2}} - t \right) - \frac{f(t)^{3/2}}{3 \sqrt{2}} = \frac{f(t)^{3/2}}{3 \sqrt{2}} - \frac{f(t)^{3/2}}{6 \sqrt{2}}
     \]
   - Simplify the right-hand side:
     \[
     \frac{f(t)^{3/2}}{3 \sqrt{2}} - \frac{f(t)^{3/2}}{6 \sqrt{2}} = \frac{2f(t)^{3/2}}{6 \sqrt{2}} = \frac{f(t)^{3/2}}{3 \sqrt{2}}
     \]

6. **Differentiate and Solve:**
   - Rearrange and differentiate the equation:
     \[
     \int_0^t f(x) \, dx + f(t) \left( \sqrt{\frac{f(t)}{2}} - t \right) - \frac{f(t)^{3/2}}{3 \sqrt{2}} = 0
     \]
   - Differentiate with respect to \(t\):
     \[
     f(t) + \frac{3}{4 \sqrt{2}} \sqrt{f(t)} f'(t) - f(t) - t f'(t) = 0
     \]
   - Simplify the equation:
     \[
     \frac{3}{4 \sqrt{2}} \sqrt{f(t)} f'(t) - t f'(t) = 0
     \]
   - Factor out \(f'(t)\):
     \[
     f'(t) \left( \frac{3 \sqrt{f(t)}}{4 \sqrt{2}} - t \right) = 0
     \]
   - Solve for \(f(t)\):
     \[
     \sqrt{f(t)} = \frac{4 \sqrt{2} t}{3}
     \]
     \[
     f(t) = \left( \frac{4 \sqrt{2} t}{3} \right)^2 = \frac{32 t^2}{9}
     \]

The final answer is \( \boxed{ f(x) = \frac{32}{9} x^2 } \).