Find the Point

Solving $x,y >0, \sqrt{x^2 + y^2} = 400, \sqrt{(x - 500)^{2} + y^{2}} + 600 = \sqrt{(x + 500)^{2} + y^{2}}$ gives $x = 240\sqrt{2}, y = 80\sqrt{7}$ . Thus, $A = 240$

Suppose $M$ is the $(x,y)$ point, $B$ is the $(-500,0)$ point, $C$ is the $(500,0)$ point and $O$ is the origin $(0,0)$ . Add the points. Draw a perpendicular $ME$ from $M$ over $BC$ . We have to find $OE$ .

$MB-MC=600$ ,

$OM=400$ ,

$BO=OC=500$

In Obtuse triangle ∆MOB , $MB^2 = MO^2+BO^2+2BO.OE$ _ _ _ (1)

In Acute triangle ∆MOC , $MC^2 = MO^2+OC^2-2OC.OE$ _ _ _ (2)

Substract (2) from (1)

$MB^2-MC^2 = BO^2-OC^2+2BO.OE+2OC.OE$

$(MB+MC)(MB-MC) = 4BO.OE$ [As BO=CO]

$(MB+MC) \times 600 = 4 \times 500 \times OE$

$OE = \frac {600 \times (MB+MC)} {2000}$ _ _ _ (3)

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From $MB-MC = 600$

$MB^2 = MC^2+1200MC+600^2$

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$MB^2+MC^2 = 2(MO^2+BO^2)$ [Apollonius theorem]

$MC^2+1200MC+600^2+MC^2=2 \times (400^2+500^2)$

$2MC^2+1200MC-460000 = 0$

$MC^2+600MC-230000 = 0$

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Now using the formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$MC = 400 \times \sqrt {2} - 300$

Similarly, $MB = 400 \sqrt {2} + 300$

$MB+MC = 800 \sqrt {2}$

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Put the value in (3)

$OE = \frac {600 \times 800 \sqrt {2} } {2000}$

$OE = x = 240 \sqrt {2}$

So $A = 240$

We have : $\sqrt{(x-500)^2+y^2} = sqrt{(x+500)^2+y^2} - 600$ and $\sqrt{x^2+y^2} = 400$ . So, now we are just need to solve the following equation to get x.

Squaring both equations, we have : $x^2+y^2 = 160000$ and $x^2-1000x+250000+y^2$ $= x^2+1000x+250000+y^2-1200\sqrt{x^2+1000x+250000+y^2}+360000$ ,

which give $1200\sqrt{410000+1000x} = 2000x+360000$ , $3\sqrt{410000+1000x} = 5x+900$

Squaring again : $3690000+9000x = 25x^2+9000x+810000$ $25x^2 = 2880000$ $x = 240\sqrt{2}$ So, we get $A = 240$

From the difference in distances we get,

sqrt( ((x+500)^2) + (y^2) )) - sqrt( ((x-500)^2) + (y^2) ) = 600 (1)

From distance from origin we get,

sqrt( (x^2) + (y^2) ) = 400 (2)

Squaring (2), we get

(x^2) + (y^2) = 160000

Simplifying (1), we get

sqrt( (x^2) + (y^2) ) + 250000 + 1000x ) - sqrt( (x^2) + (y^2) + 250000 - 1000x ) = 600

Replacing (x^2) + (y^2) = 160000 and squaring gets,

410000 + 1000x + 410000 - 1000x - 2*( sqrt( (410000^2) - 1000000(x^2) ) ) = 360000

This gives, 230000 = sqrt( (410000^2) - 1000000(x^2) )

Squaring on both sides gets, (410000^2) - (230000^2) = 1000000(x^2)

(x^2) = 1152/100

Or x = 240*sqrt(2)

Thus A = 240

In Analytic Geometry, the distance between two points a & b can be calculated using the following formula :

ab = \sqrt{(x 2-x 1)^2 + (y 2-y 1)^2} ,where a(x 1, y 1), b(x 2, y 2)

So, using this and the problem information, the following is true :

\sqrt{(x-500)^2 + y^2} + 600 = \sqrt{(x-(-500))^2 + y^2} \sqrt{x^2 - 1000x + 250000 + y^2} + 600 = \sqrt{x^2 + 1000x + 250000 + y^2} \rightarrow 1 \sqrt{x^2 + y^2} = 400 Then x^2 + y^2 = 160000 y^2 = 160000 - x^2 \rightarrow 2

By using y^2 from the second equation in the first equation, Then

\sqrt{x^2-1000x+250000+160000-x^2}+600 = \sqrt{x^2+1000x+250000+160000-x^2} then \sqrt{410000+1000x}-\sqrt{410000-1000x} = 600 \sqrt{4100+10x}-\sqrt{4100-10x} = 60 4100+10x+4100-10x-2 \sqrt{4100+10x} \sqrt{4100-10x} = 60 4100-\sqrt{4100+10x} \sqrt{4100-10x} = 1800 \sqrt{16810000-100x^2} = 2300 16810000-100x^2 = 5290000 x^2 = 168100-52900 = 115200 x = \sqrt{2} 240 = \sqrt{2}*A Then A = 240

From the second condition, we get that $\sqrt{x^2+y^2}=400$ or $x^2+y^2=160000$

From the first condition, we get that $\sqrt{(x-500)^2+y^2}-\sqrt{(x+500)^2+y^2}=600$ $\sqrt{x^2-1000x+250000+y^2}-\sqrt{x^2+1000x+250000+y^2}=600$ substituting the first equation we found in, $\sqrt{-1000x+250000+160000}-\sqrt{1000x+250000+160000}=600$ $\sqrt{-1000x+410000}-\sqrt{1000x+410000}=600$ squaring both sides gives, $-1000x+410000+1000x+410000-2\sqrt{168100000000-1000000x^2}=360000$ (820000-2\sqrt{168100000000-1000000x^2}=360000) $-2\sqrt{168100000000-1000000x^2}=-460000$ $\sqrt{168100000000-1000000x^2}=230000$ squaring again gives, $168100000000-1000000x^2=52900000000$ $1000000x^2=115200000000$ $x^2=115200$ so $x=240sqrt{2}$ and the $A=\boxed{240}$

Distance from origin to (x,y) is 400 so x^2+y^2=(400)^2 the question, mathematically, states that root((x+500)^2+y^2) + 600 = root((x-500)^2+y^2) =>root(400^2+500^2+1000x) + 600 = root(400^2+500^2-1000x) now, putting k^2=400^2+500^2 we get root(k^2+1000x) - root(k^2-1000x) = -600 squaring, 2k^2-2root(k^4-10^6x^2)=600^2 putting k^2 value and x^2=2A^2, 2A^2.10^6=(41^2-23^2).10^8 =>A=8.3.10=240

Since (x,y) is in first quadrant and is 600 units closer to (500,0) than (-500,0), then by using distance formula: $\sqrt{(x+500)^2 + y^2} = \sqrt{(x-500)^2 + y^2} + 600$ and (x,y) is 400 from origin, hence: $\sqrt{x^2 + y^2} = 400$ and $x^2 + y^2 = 400^2$ after substituting $y^2 = 400^2 - x^2$ into the first equation and further calculation for $x^2$ , we'll get $x^2 =115200$ and $x = 240\sqrt{2}$ .Hence, $A=240$

lets start with "a point (x,y) is a distance of 400 units away from the origin."

using distance between two points formula : \sqrt { (y2-y1)^2 + (x2-x1)^2 }

we get : \sqrt{(x-0)^2 + (y-0)^2)} = 400 removing sqrt we get x^2 + y^2 = 160000

so y^2 = 160000 - x^2 \rightarrow equ 1

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assuming distance between point (x,y) and point (-500,0) is "z" using same formula:

z= \sqrt{(x+500)^2 + (y-0)^2} z^2= x^2 +1000x +250000 +y^2

eliminating y^2 by substituting from equ 1

we get :

z^2= 410000 +1000x \rightarrow equ 2

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as " point (x,y) is 600 units closer to the point (500,0) than to the point (−500,0)" and if distance between point (x,y) and point (-500,0) is "z" then distance between point (x,y) and (500,0) must be 600 unit smaller . so distance between point(x,y) and point (500,0) equals "z-600" .

using same formula:

z-600 = \sqrt{(x-500)^2 + (y-0)^2} removing sqrt we get: (z-600)^2=(x-500)^2+y^2

z^2-1200z+360000=x^2-1000x+250000+y^2 z^2-1200z=x^2-1000x+y^2 - 110000 eliminating y^2 by substituting from equ 1

we get:

z^2-1200z=50000-1000x \rightarrow equ 3

by substituting with equ 2 in equ 3 z^2= 410000 +1000x \rightarrow equ 2 z=\sqr{410000 +1000x}

z^2-1200z=50000-1000x \rightarrow equ 3

we get:

410000 +1000x-1200(\sqr{410000 +1000x})=50000-1000x 2000x+360000-1200(\sqr{410000 +1000x})=0 2000x+360000=1200(\sqr{410000 +1000x}) 20x+3600=12(\sqr{410000 +1000x})

removing sqrt:

400x^2+(3600)^2+144000x=144(410000 +1000x) with some math 400x^2+(3600)^2+144000x=144000x+59040000

400x^2=46080000

so x^2=115200

given that x = \sqrt{2} \times A

\sqrt{115200}= \sqrt{2} \times A \frac {\sqrt{115200}}{\sqrt{2}} = A \sqr{\frac {115200}{2}} =A \sqr{57600}=A

so A =240

First quadrant means both x and y are positive,

First condition:

√(〖(x+500)〗^2+y^2 )-√(〖(x-500)〗^2+y^2 )=600----(i)

Second condition:

x^2 + y^2 =〖400〗^2---(ii)

From (i)

√(x^2+1000x+〖500〗^2+y^2 )+√(x^2-1000x+〖500〗^2+y^2 )=600

√(〖400〗^2+1000x+〖500〗^2 )+√(〖400〗^2-1000x+〖500〗^2 )=600-----[from (ii)---> x^2+y^2=〖400〗^2]

Squaring both sides

2 (〖400〗^2+〖500〗^2 )-2 √(〖(〖400〗^2+〖500〗^2)〗^2-〖(1000*x)〗^2 )=〖600〗^2

2 (〖400〗^2+〖500〗^2 )-〖600〗^2 = 2 √(〖(〖400〗^2+〖500〗^2)〗^2-〖(1000*x)〗^2 )

√(〖(〖400〗^2+〖500〗^2)〗^2-〖(1000*x)〗^2 ) = 230000

(1000*x)〗^2=〖〖〖(400〗^2+〖500〗^2)〗^2-230000〗^2

x =339.411255

√2 A=339.411255

A = 240

The assertion that $(x,y)$ is $600$ units closer to $(500,0)$ than to $(-500,0)$ means that $(x,y)$ is a point on the hyperbola $\frac{x^2}{300^2} - \frac{y^2}{500^2 - 300^2} = 1$ , subject to $x>0$ . Note that $500^2 - 300^2 = 400^2$ , so this equation can be simplified to $\frac{x^2}{300^2} - \frac{y^2}{400^2} = 1$ .

Additionally, since $(x,y)$ is $400$ units from the origin, it must be a point on the circle $x^2 + y^2 = 400^2$ . Therefore, $\frac{y^2}{400^2} = 1 - \frac{x^2}{400^2}$ , so we can substitute this expression for $\frac{y^2}{400^2}$ into the equation for the hyperbola and get $\frac{x^2}{300^2} - 1 + \frac{x^2}{400^2} = 1$ . Hence, $x^2( 400^2 + 300^2) = 2 (400^2) (300^2)$ and $x = \frac{\sqrt{2}\cdot 400 \cdot 300}{500} = 240 \sqrt{2}$ . Thus, the value of $A$ we are looking for is $240$ .