Find the point P on the parabola y=(x^2) closest to the point (3,0).?

the clear answer in guide claims that it’s (0.59,0.35).

The exact distance between a place (x, y) regarding parabola while the point (3, 0) is provided by sqr((x – 3)^2 + (y – 0)^2) (in which sqr implies square-root). You want to reduce this.
We could disregard the square-root and simply reduce (x – 3)^2 + y^2 (the square associated with length). And because y = x^2, we are able to replace and reduce (x – 3)^2 + x^4. Therefore we wish get a hold of at the least
f(x) = x^4 + x^2 – 6x + 9.
Initial by-product is provided by
f'(x) = 4x^3 + 2x – 6.
Establishing this corresponding to zero, we need to resolve:
4x^3 + 2x – 6 = 0.
By assessment we are able to note that one option would be x = 1. Officially we have to validate that should indeed be the very least, nonetheless it seems like the idea (1, 1) could become the main one we wish.
Examining the 2nd by-product:
f”(x) = 12x^2 + 2
This will be good for x = 1, therefore we’ve got a nearby minimal at x = 1.
Indeed it’ll become a complete minimal besides, because various other two solutions of f'(x) = 0 will become imaginary. But i will not show that right here. (aspect completely x – 1 and appear in the quadratic phrase that is remaining.)
EDIT: Heh, we see Puggy got a post with the perfect solution is while I happened to be composing my own. And Puggy’s edit is right-about the perfect solution is provided within guide. Have you been yes you are looking in the response to the proper issue?

y = x^2 + tx – 2 y = 3x + 3 intersects the parabola at facets P and R, therefore from looking around regarding drawing, the range’s x-intercept is factor P: 0 = 3x + 3 => P = (-a million, 0) today plugging factor P to the special y equation: 0 = a million – t – 2 => t = -a million placing y = 0, we might have the ability to find out factor Q(different x-intercept). x^2 – x – 2 = 0 => (x – 2)(x + a million) = 0 Q = (2, 0). allow factor roentgen be written as (x, x^2 – x – 2), it uses from range y = 3x + 3 that: (x^2 – x – 2 – 0)/(x + a million) = 3 (utilizing the equation for pitch) x^2 – x – 2 = 3x + 3 x^2 – 4x – 5 = 0 (x – 5)(x + a million) = 0 x can fundamentally comparable 5 because that factor roentgen is within quadrant a million the positioning x > 0. roentgen = (5, 5^2 – 5 – 2) = (5, 18) part of triangle PQR = (difference in x-intercepts)(difference in y-values between facets Q and R)/2 = (3)(18)/2 = 27

Goal: to reduce the exact distance between (x, y) and (3, 0), regarding bend y = x^2.
What’s the length between (x, y) and (3, 0)? We determine this utilizing the length formula.
D = sqrt( (x2 – x1)^2 + (y2 – y1)^2 )
Allow (x1, y1) = (3, 0) and (x2, y2) = (x, y). After That
D = sqrt( (x – 3)^2 + (y – 0)^2 )
D = sqrt( (x – 3)^2 + y^2 )
Understand that, since y = x^2, we are able to change every example of y for the reason that formula with x^2.
D = sqrt( (x – 3)^2 + (x^2)^2 )
D = sqrt( (x – 3)^2 + x^4 )
You want to reduce this by simply making dD/dx = 0.
We *could* differentiate this right, however when we need to cope with types with origins followed closely by the string guideline. I’ll alternatively square both edges, after which differentiate implicitly.
D^2 = (x – 3)^2 + x^4
Differentiate implicitly pertaining to x,
2D (dD/dx) = 2(x – 3) + 4x^3
Make dD/dx = 0, getting
0 = 2(x – 3) + 4x^3
Resolve for x.
0 = 2x – 6 + 4x^3
0 = 4x^3 + 2x – 6
0 = 2x^3 + x – 3
We’ve a cubic that individuals want to factor. Luckily for people, if p(x) = 2x^3 + x – 3, p(1) = 0, which informs us straight away (x – 1) is an issue. This implies we are able to have the various other facets by artificial lengthy unit.
Missing the measures of the, our cubic facets like therefore:
0 = (x – 1)(2x^2 + 2x + 3)
2x^2 + 2x + 3 does not factor, and it is constantly good we are able to successfully dismiss that component due to that.
For that reason, x = 1, and our nearest point is present at x = 1.
But we wish the y-coordinate also.
Since y = x^2, after that y = 1^2 = 1,
therefore the nearest point is (1, 1).
Why don’t we compare it because of the various other responses by determining their particular length:
1) length between (3, 0) and (1, 1):
D = sqrt( (3 – 1)^2 + (0 – 1)^2 )
= sqrt( 2^2 + (-1)^2 )
= sqrt(4 + 1)
= sqrt(5)
2) length between (3, 0) and (2, 4):
D = sqrt ( (3 – 2)^2 + (0 – 4)^2 )
= sqrt( 1^2 + (-4)^2 )
= sqrt(1 + 16)
= sqrt(17) [fails; sqrt(5) < sqrt(17) ) 3) Distance between (3, 0) and (1.29, 1.66) (going to use approximations since the difference is so miniscule) D ~= sqrt( (3 - 1.29)^2 + (1.66 - 4)^2 ) D ~= sqrt( (1.71)^2 + (5.4756) ) D ~= sqrt( 2.9241 + 5.4756 ) D ~= 2.9 but sqrt(5) ~= 2.23 so sqrt(5) < 2.9, making #1 distance smaller than #2 and #3. **EDIT** Let's try out the answer you've given, which doesn't make sense since I worked it out algebraically. If your answer is truly the correct answer, your answer's distance should be less than sqrt(5). 4) Distance between (3, 0) and (0.59, 0.35). Hmm... your question already fails, because (0.59, 0.35) is not a point on the curve. Since y = x^2, it follows that squaring 0.59 should give 0.35 (it doesn't).

Distance from (3,0) to the parabola is
D=sqrt[(x-3)^2+(y-0)^2]
D=sqrt[ x^2-6x+9 + y^2]
D=sqrt[ x^2-6x+9 +x^4]
dD/dx = (2x-6+4x^3)/2sqrt[x^2-6x+9 +x^4]=0
-4x^3+2x-6=0
4x^3-2x-6=0
Resolve this equation for x graphically or utilizing a calculator.
x=1.28962390
y=x^2 = 1.66312980345121
(1.2896, 1.6631) will be the nearest things
It should be confirmed that d^2D/dx^2 > 0 whenever x=1.2896

The guide should be incorrect. (0.59,0.35) is a place regarding parabola, however it is perhaps not the nearest point out (3,0).
oramirez012

Answer 6

its a challenging concern however, if you intend to resolve it you can easily draw a perpendicular from (3,0) to a tangent associated with parabola.
after which it may be simple to find it.

Answer 7

I am uncertain but I believe it is (2,4)

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