? asked in Science & MathematicsMathematics · 10 years ago

Distance from curve to origin in R3.?

Find the minimum distance from the origin to the curve 3x^2 + 4xy +3y^2 = 20

A) 1

B) 3sqrt(2)

C) 2

D) 23sqrt(2)

E) 53sqrt(2)

It has been so long since I have done this and I am trying to help my nephew you out but I have completely forgot can someone help

1 Answer

  • S
    Lv 7
    10 years ago
    Favorite Answer

    this is so weird; I read this problem and putzed with it and could not solve.

    However, I woke up this morning with the answer!!

    look at both the origin and the form of your equation. Let's say you were blindfolded and someone came along and replaced every x with a y and every y with an x (***including the origin*** - key point) and reshowed you the problem. Ta da!! It's the same problem. so.... your answer has to be of the from x=y (also works if x-->-y and y--->-x)

    replace every y and x with X

    3X^2 + 4X^2 +3X^2 =20

    10 X^2 = 20

    X^2 =2

    distance is sqrt (x^2 +y^2) = sqrt(4)= 2

    answer C

    ta da!!!!!!!! damn I'm good!


    let's do a check and see if c works (sqrt(2), sqrt(2)) OR

    (-sqrt(2), -sqrt(2) )

    3*2 +4*2 +3*2 =20

    6+8+6 =20 yep it works

    therefore 2 is A distance from (0,0) to the curve (specifically at the points (sqrt(2), sqrt(2)) OR (-sqrt(2), -sqrt(2) )

    that ELIMINATES B,D,E as they are all greater than 2.

    So the only 2 possible answers are A or C on the basis of "guessing" that ( Sqrt(2), sqrt(2) ) is a point on the curve (it is) that is close to 0,0. But we had a basis for choosing that point as detailed above. It's C on the basis of the symmetry of the equation.

    But just to be fanatical with this....I showed that c (2) is a valid distance and since we are looking a minimum distance we can eliminate b,d,e. But just in case you don't believe my explanation - let's just look at a = 1 and eliminate it.

    pretend that a is correct (distance is 1); that implies the answer lies on this circle: x^2+y^2 =1

    we can write your equation as 3(x^2+y^2) +4xy =20

    BUT the stuff in the brackets is 1.

    so....3*1 +4xy = 20

    xy = 17/4

    but I can parameterize my original circle by x= cos(a) and y = sin(a) and rewrite it as cos^2(a) + sin^2(a) = 1 which is a trig identify. In other words, we are looking for a magic angle (a) that does the trick.

    so xy =17/4 can be written as

    cos(a)*sin(a) = 17/4

    there is no a that solves this equation; the reason is the aboslute value of cos(a) and sin(a) by themselves is 1

    so cos(a)*sin(a) <1 and cannot reach 17/4

    therefore the curve lies further than 1 unit away and answer A is eliminated.

    Source(s): PhD Biochemist that took a lot of math - specifically including group theory that utilizes symmetry. Your problem is symmetric in x & y.
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