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

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.)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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