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Dear Aditya,
its distance = sqrt ( (3-x)^2 + (2-y)^2 ) right?So since y = x² + 2, you can substitute this into the above so that you have a function of only x, no y in itd = sqrt ( (3-x)^2 + (2 - x² - 2)^2 )Now, how do you minimize d? I assume you know how to do differential calculus because you should for this questionthe first derivative is 1/2*(-6+2*x+4*x^3)/sqrt(9 - 6*x + x^2+ x^4) you set it to 0 and solve for x, You get 3 roots, one of which is real, the other are complex. Discard the complex roots, and take x = 1, which will satisfy the derivative = 0Substitute x = 1 into d = sqrt ( (3-x)^2 + (2 - x² - 2)^2 ),and you get d = sqrt(5) or 2.236067977in fact x = 1, y = 3, is the point on the curve closest to the point (3, 2)
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