Solve Shortest Distance Point (0,c) from Parabola y=x^2

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SUMMARY

The discussion focuses on finding the shortest distance from the point (0, c) to the parabola defined by the equation y = x^2, specifically for values of c within the range 0 ≤ c ≤ 5. The solution provided, √(c - 0.25), is valid only when c > 1/4. For values of c less than or equal to 1/4, the nearest point on the parabola must be evaluated at the endpoints of the interval, indicating that the solution requires careful consideration of the entire range of c.

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  • Understanding of calculus concepts, particularly optimization.
  • Familiarity with the properties of parabolas and their equations.
  • Knowledge of distance formulas in a Cartesian coordinate system.
  • Basic understanding of interval notation and its implications in mathematical problems.
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  • Study optimization techniques in calculus to solve similar distance problems.
  • Learn about the geometric properties of parabolas and their applications.
  • Explore distance minimization problems in coordinate geometry.
  • Investigate the implications of boundary conditions in mathematical analysis.
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gaganpreetsingh
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Hi I am stuck on this ques. although i have solved and got the ans but i am still facing a problem.
Find the shortest distance of the point (0,c) from the parabola y=x^2 where 0<= c <= 5
i got the ans right that is
(c-0.25)^0.5
but i have not been able to understand the use of the 0<= c <= 5
any help?
 
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any help out there?
 
Assuming that c> 1/4, I get that. However, notice that your solution makes no sense if c< 1/4! What is the nearest point on the parabola if c< 1/4? (Hint: check the endpoints of the interval of values for y.)
 

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