closest-distance-point-3-0-to-curve-y-2-x-4-

Question Number 83495 by john santu last updated on 03/Mar/20

closest distance point (3, 0) to curve

y^{2} = x + 4 ?

Commented by john santu last updated on 03/Mar/20

d = \sqrt{{(x - 3)}^{2} + {(y)}^{2}}

d^{2} = {(x - 3)}^{2} + y^{2}

x + 4 + x^{2} - 6 x + 9 - d^{2} = 0

x^{2} - 5 x + 13 - d^{2} = 0

tangency △ = 25 - 4 (13 - d^{2}) = 0

\frac{25}{4} = 13 - d^{2} \Rightarrow d = \sqrt{\frac{52 - 27}{4}} = \frac{3 \sqrt{3}}{2}

Commented by john santu last updated on 03/Mar/20

dear Mr W . i try your method .

this correct ?

Commented by jagoll last updated on 03/Mar/20

good sir

Commented by jagoll last updated on 03/Mar/20

x^{2} - 5 x + 13 - d^{2} = 0

tangency △ = 25 - 4 (13 - d^{2}) = 0

13 - d^{2} = \frac{25}{4} \Leftrightarrow d^{2} = \frac{52 - 25}{4}

d = \sqrt{\frac{27}{4}} = \frac{3 \sqrt{3}}{2}

Commented by john santu last updated on 03/Mar/20

oo yes sir . thank you