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The smallest distance between the origin and a point on the parabola $y=x^2-5$ can be expressed as $\sqrt{a}/b$, where $a$ is not divisible by the square of any prime. Find $a+b$.

michaelcai  Aug 11, 2017
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#1
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Let  u be the x coordinate of the point we seek....so u^2  - 5 is the y coordinate

And we are  seeking to minimize tthe distance, D, expressed as

D  = [ (u - 0)^2 + ( u^2  - 5 -  0 )^2 ]^(1/2)

D  = [ u^2  + u^4 - 10u^2  + 25 ]^(1/2)

D  = [ u^4  - 9u^2 + 25 ]^(1/2)

Take the derivative of this  and set it to 0

D'  = (1/2)   [ u^4  - 9u^2 + 25 ]^(-1/2) * [ 4u^3 - 18u]  = 0

This will = 0 whenever [ 4u^3 - 18u]  = 0

Factor

u [ 4u^2  - 18]  = 0

Setting both factors to 0

Either  u = 0  and  u^2 - 5  = 5    ....so one possible point is  (0, -5)  and this is 5 units from the origin

Or

4u^2  - 18  = 0

4u^2  = 18

u^2  = 18/4

u  = ± √ [18 /4]  = ± 3√2 / 2      and  u^2  - 5  =  18/4 - 5 =  18/4 - 20 / 4  = - 2/4 =   (-1/2)

So....the other points are  ( 3√2 / 4 , - 1/2 )   and  ( - 3√2 / 4 , - 1/2 )

And the distance that either one of these points is from the origin is given by

D  = √  [  (18/4)^2 - 9 (18/4)  + 25 ]  =  √  [ 324 - 648 + 400] / 4  =  √76 / 4  = 2√19/ 4  =  √19/2 ≈ 2.179

And this is the minimum distance from the origin to any point on the parabola

So...... √a / b  =    √19 / 2      and   a +  b  =  19  + 2  = 21

CPhill  Aug 11, 2017
edited by CPhill  Aug 12, 2017

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